# Limits and Continuity

This section covers:

Note that we discuss finding limits using L’Hopital’s Rule here.

# Introduction to Limits

We need to understand how limits work, since the first part of Differential Calculus (calculus having to do with rates at which quantities change) uses them. I like to think of a limit as what the $$y$$ part of a graph or function approaches as $$x$$ gets closer and closer to a number, either from the left-hand side (which means that $$x$$ part is increasing), or from the right hand side (which means the $$x$$ part is decreasing).

We can write a limit where $$x$$ gets closer and closer to 0 as $$\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)=L$$. To describe this, we say the “limit of $$f\left( x \right)$$ as $$\boldsymbol{x}$$ approaches 0 is $$\boldsymbol{L}$$”. Now the beauty of limits is that $$x$$ can get closer and closer to a number, but not actually ever get there (think asymptote from the Rational Functions sections). Even if a function is “normal”, like a linear function, we still consider the $$y$$-value a limit, as shown below.

The reason we have limits in Differential Calculus is because sometimes we need to know what happens to a function when the $$x$$ gets closer and closer to a number (but doesn’t actually get there); we will use this concept in getting the approximation of a slope (“rate”) of a curve at that point. Sometimes, the $$x$$ value does get there (like when we’re taking the slope of a straight line), but sometimes it doesn’t (like when we’re taking the slope of a curved function).

As an example, when you first learn how to handle limits, it might be the case that the $$x$$ value is getting closer and closer to a number that makes the denominator of the $$y$$ value 0; this can’t happen, or the fraction will “blow up”.

Here’s a graphical example of a removable discontinuity, or hole, that represents a limit at the $$x$$ value where the discontinuity exists. Note that $$y$$ is undefined at $$x=1$$, but the limit at $$x=1$$ is defined (it is 3, what would have been the $$y$$-value without the hole). (Note that we had to factor a (difference of cubes) in this example).

Limit Graph

$$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\frac{{{{x}^{3}}-1}}{{x-1}}=3$$

T-Chart:

 x y –1 1 0 1 .9 2.71 .95 2.8525 .99 2.9701 1 Error 1.01 3.0301 1.05 3.1525 1.1 3.31
Note that $$\require{cancel} \displaystyle \frac{{{{x}^{3}}-1}}{{x-1}}=\frac{{\cancel{{\left( {x-1} \right)}}\left( {{{x}^{2}}+x+1} \right)}}{{\cancel{{x-1}}}}={{x}^{2}}+x+1$$, so we have a quadratic with a removable discontinuity at $$x=1$$:

Notice how as $$x$$ gets closer and closer to 1, $$y$$ gets closer and closer to 3!

So, $$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\frac{{{{x}^{3}}-1}}{{x-1}}=3$$.

# Finding Limits Algebraically

We will learn different techniques for finding simple limits; here are some (non-trig) problems:

 Rule Problem Solution Simply Plugging in $$x$$ Value $$\underset{{x\to 4}}{\mathop{{\lim }}}\,\left( {3-x} \right)$$ I know it seems obvious, but if we can simply plug in the limit and the function is still defined; go for it! $$\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,\left( {3-x} \right)=3-4=-1$$ Simplifying (such as Factoring, Combining Terms) $$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\frac{{{{x}^{3}}-1}}{{x-1}}$$         $$\displaystyle \underset{{x\to 6}}{\mathop{{\lim }}}\,\frac{{\frac{1}{3}-\frac{1}{{x-3}}}}{{x-6}}$$ We saw this first example in the graph above; we can factor the top (difference of cubes),  simplify, and then plug in the 1 in the resulting expression: $$\displaystyle \frac{{{{x}^{3}}-1}}{{x-1}}=\frac{{\cancel{{\left( {x-1} \right)}}\left( {{{x}^{2}}+x+1} \right)}}{{\cancel{{x-1}}}}={{x}^{2}}+x+1$$ $$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\frac{{{{x}^{3}}-1}}{{x-1}}=\underset{{x\to 1}}{\mathop{{\lim }}}\left( {{{x}^{2}}+x+1} \right)={{\left( 1 \right)}^{2}}+1+1=3$$   In the second example, we can find a common denominator and combine terms: $$\displaystyle \frac{{\frac{1}{3}-\frac{1}{{x-3}}}}{{x-6}}=\,\frac{{\frac{{x-3}}{{3\left( {x-3} \right)}}-\frac{{1\cdot 3}}{{3\left( {x-3} \right)}}}}{{x-6}}=\frac{{\frac{{x-6}}{{3\left( {x-3} \right)}}}}{{x-6}}=\frac{{\cancel{{x-6}}}}{{3\left( {x-3} \right)\cancel{{\left( {x-6} \right)}}}}=\frac{1}{{3\left( {x-3} \right)}}$$ $$\displaystyle \underset{{x\to 6}}{\mathop{{\lim }}}\,\frac{1}{{3\left( {x-3} \right)}}=\frac{1}{9}$$ Rationalizing the Fraction (typically if you see a fraction with a square root in it) $$\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\frac{{\sqrt{{x+11}}-3}}{{x+2}}$$ We can see that we can’t directly plug in –2, since the denominator will be 0. There is a trick where we can rationalize the numerator by multiplying by a fraction with the conjugate (you change the sign in the middle of the two terms) on top and bottom. Notice we use difference of squares to simplify: \displaystyle \begin{align}\frac{{\sqrt{{x+11}}-3}}{{x+2}}&=\left( {\frac{{\sqrt{{x+11}}-3}}{{x+2}}} \right)\left( {\frac{{\sqrt{{x+11}}+3}}{{\sqrt{{x+11}}+3}}} \right)\\&=\frac{{{{{\left( {\sqrt{{x+11}}} \right)}}^{2}}-{{3}^{2}}}}{{\left( {x+2} \right)\left( {\sqrt{{x+11}}+3} \right)}}=\frac{{x+11-9}}{{\left( {x+2} \right)\left( {\sqrt{{x+11}}+3} \right)}}\\&=\frac{{\cancel{{x+2}}}}{{\cancel{{\left( {x+2} \right)}}\left( {\sqrt{{x+11}}+3} \right)}}=\frac{1}{{\sqrt{{x+11}}+3}}\end{align}   Now, let’s evaluate the limit: $$\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{\sqrt{{x+11}}-3}}{{x+2}}=\underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{1}{{\sqrt{{x+11}}+3}}=\frac{1}{{\sqrt{{\left( {-2} \right)+11}}+3}}=\frac{1}{6}$$ Putting in a number really close in your calculator $$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}$$ If all else fails and you can use your calculator, try putting in a number really close to the $$x$$ value; for example, put in .99999 or  1.00001 or for 1: You can see that the limit is 3. Later, we’ll see that it may matter if we pick a number a little smaller or larger, depending on which side of the $$x$$ we are coming from.

Again, we discuss finding limits using L’Hopital’s Rule here.

# Continuity and One Side Limits

Sometimes, the limit of a function at a particular point and the actual value of that function at the point can be two different things. Notice in cases like these, we can easily define a piecewise function to model this situation.

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The limit from the right, or $$\underset{{x\to {{c}^{+}}}}{\mathop{{\lim }}}f\left( x \right)=L$$ means that $$x$$ approaches $$c$$ from the right side, or with values greater than $$c$$, and the limit from the left, or $$\underset{{x\to {{c}^{-}}}}{\mathop{{\lim }}}f\left( x \right)=L$$ means that $$x$$ approaches $$c$$ from the left side, or with values less than $$c$$. Do you see how if the limit from the right and the limit from the left are the same, then we can get a “regular” limit (meaning both sides converge to the same $$y$$ value?)

## Existence of a Limit and Definition of Continuity

Do you also see that if the limit from the right equals the limit from the left, and this equals the actual point for $$f\left( x \right)$$ (the $$y$$ for that $$x$$), then we have a continuous function (one that we can draw without picking up our pencil)? This leads to the definition of the existence of a limit, the formal definition of continuity:

 Limit Theorem/Definition Explanation Existence of a Limit   $$\underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)=L$$  if and only if   $$\underset{{x\to {{c}^{-}}}}{\mathop{{\lim }}}\left( x \right)=L$$  and   $$\underset{{x\to {{c}^{+}}}}{\mathop{{\lim }}}\left( x \right)=L$$ What this says is that if you don’t get closer and closer to a number (the “$$y$$” or $$f\left( x \right)$$) from both sides of the “$$x$$”, then there is no limit of  at that point.   The actual point $$f\left( c \right)$$ may be defined (as in a non-continuous function) at a completely different $$y$$ (where no limit may occur), but in order for a limit to occur, the $$x$$’s have to approach a certain $$y$$ value from both sides.   As an example, note that there would be an existence of a limit at a removable discontinuity  (“hole”) in a function, but not at a vertical asymptote, unless the function is approaching the vertical asymptote in the same direction $$(-\infty \,\,\text{or}\,\,\infty )$$ . Definition of Continuity   A function is continuous at $$c$$ when the following three conditions are met:   $$f\left( c \right)$$ is defined $$\underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)$$ exists $$\underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)=f\left( c \right)$$ Basically, I just like to think of continuity as whether or not you have to “pick up your pencil” when you draw a function from left to right.   The definition makes sense, however, since if the $$x$$’s from both directions are getting closer and closer to a number (the “$$y$$”), and the actual point is that number, then you don’t really need to pick up your pencil to draw the function.

Here are some examples; remember that the actual limits are the $$y$$ values, not the $$x$$. The first example shows that some limits do not exist (DNE), based on the definition above. The second example actually gives you the equation for the piecewise function that illustrates limits. Notice that both functions are discontinuous.

 Graphs Limits and Points \begin{align}\underset{{x\to -{{1}^{-}}}}{\mathop{{\lim }}}\,f\left( x \right)&=0\,\\\underset{{x\to -{{1}^{+}}}}{\mathop{{\lim }}}\,f\left( x \right)&=-2\\\underset{{x\to -1}}{\mathop{{\lim }}}\,f\left( x \right)&=\text{DNE}\\f\left( {-1} \right)&=-2\end{align}           \begin{align}\,\underset{{x\to {{1}^{-}}}}{\mathop{{\lim }}}\,f\left( x \right)&=4\\\underset{{x\to {{1}^{+}}}}{\mathop{{\lim }}}\,f\left( x \right)&=-\infty \\\underset{{x\to 1}}{\mathop{{\lim }}}\,f\left( x \right)&=\text{DNE}\\\,f\left( 1 \right)&=0\end{align}   DNE = does not exist \begin{align}\underset{{x\to -{{1}^{-}}}}{\mathop{{\lim }}}\,f\left( x \right)&=-4\,\\\underset{{x\to -{{1}^{+}}}}{\mathop{{\lim }}}\,f\left( x \right)&=-4\\\underset{{x\to -1}}{\mathop{{\lim }}}\,f\left( x \right)&=-4\\f\left( {-1} \right)&=-2\end{align}

# Continuity of Functions

We learned in the Graphing Rational Functions, including Asymptotes section how to find removable discontinuities (holes) and asymptotes of functions (basically anywhere where we’d get a 0 in the denominator of the function); now we know that these functions are discontinuous at these points.

Let’s review how we get vertical asymptotes for a rational function:

 Holes and Vertical Asymptotes Examples First factor both the numerator and denominator, and cross out any factors in both the numerator and denominator.   ⇒ If any of these factors contain variables, these are removable discontinuities, or “holes” and will be little circles on the graphs. The idea is that if you cross out a polynomial, you can’t forget that it was in the denominator and can’t “legally” be set to 0.   ⇒ The domain of a rational function is all real numbers, except those that make the denominator equal 0, as we saw earlier.   ⇒ (Note that if after you cross out factors, you still have that same factor on the bottom, the “hole” will turn into a vertical asymptote; follow the rules below). $$\require{cancel} \displaystyle y=\frac{{{{x}^{2}}-5x+6}}{{x-3}}=\frac{{\cancel{{\left( {x-3} \right)}}\left( {x-2} \right)}}{{\cancel{{\left( {x-3} \right)}}}}=x-2$$   This function reduces to the line $$y=x-2$$ with a removable discontinuity (a little circle on the graph) where $$x=3$$ and $$y=\left( 3 \right)-2=1$$ (plug in 3 for $$y$$ in reduced fraction). So, the hole is at $$(3,1)$$. (Note that you can see this hole on your TI graphing calculator with “zoom 4”  or “zoom ZDecimal”).   Domain is $$\left( {-\infty ,3} \right)\cup \left( {3,\infty } \right)$$, since a 3 would make the denominator $$=0$$. It’s like we have to “skip over” the 3 with interval notation. To get vertical asymptotes or VAs:   ⇒ After determining if there are any holes in the graph, factor (if necessary) what’s left in the denominator and set the factors to 0.  For any value of $$x$$ where these factors could be 0, this creates a vertical asymptote at “$$x=$$” for these values.   ⇒ Note: There could a multiple number of vertical asymptotes, or no vertical asymptotes.   ⇒ Don’t forget to include the factors with “$$x$$” alone ($$x=0$$ is the vertical asymptote). $$\displaystyle y=\frac{{{{x}^{2}}-5x+6}}{{x\left( {{{x}^{2}}-9} \right)}}=\frac{{\cancel{{\left( {x-3} \right)}}\left( {x-2} \right)}}{{x\cancel{{\left( {x-3} \right)}}\left( {x+3} \right)}}=\frac{{x-2}}{{x\left( {x+3} \right)}}$$   Vertical asymptotes occur when $$\left( {x-0} \right)=0$$ or $$\left( {x+3} \right)=0$$, or $$x=0,-3$$.   Domain is $$\left( {-\infty ,-3} \right)\cup \left( {-3,0} \right)\cup \left( {0,3} \right)\cup \left( {3,\infty } \right)$$, since anything that could make the denominator 0 (even a hole) can’t be included. So we have to “skip over”  –3, 0, and 3.

And we’ll also have to remember the trig function asymptotes:

 Trig Function Asymptote Sine (Sin),  Cosine (Cos) None Tangent (Tan),  Secant (Sec) $$\displaystyle \frac{\pi }{2}\,\,+\,\,\pi k$$ Cosecant (Csc),  Cotangent (Cot) $$\pi k$$

In Calculus, you may be asked to find the $$x$$-values at which a function might be discontinuous, and also determine whether or not a discontinuity is removable or non-removable:

 Function Continuous or Discontinuous? $$\displaystyle f\left( x \right)=\frac{x}{{{{x}^{2}}+2x}}$$ We need to factor the denominator and then we can see that we have one removable discontinuity at $$x=0$$, and one non-removable discontinuity (vertical asymptote), $$x=-2$$. (Remember that factors that can be removed result in removable discontinuities, or holes.) $$\require{cancel} \displaystyle f\left( x \right)=\frac{x}{{{{x}^{2}}+2x}};\,\,\,\,f\left( x \right)=\frac{{\cancel{x}}}{{\cancel{x}\left( {x+2} \right)}}=\frac{1}{{\left( {x+2} \right)}}$$ But notice that we can just see where denominator $$=0$$ to see where the function is discontinuous: at $$x=0$$ and $$x=-2$$. $$\displaystyle f\left( x \right)=\frac{{\left| x \right|}}{x}$$ Let’s first turn this into a piecewise function, with the “boundary point” at $$x=0$$ (we set what’s inside the absolute value to 0): $$\displaystyle f\left( x \right)=\left\{ \begin{array}{l}\,\,\,\,1\text{; }\,\,\,\,\,x>0\\-1\text{;}\,\,\text{ }\,\,\text{ }x<0\end{array} \right.$$ But at $$x=0$$, the function is undefined since we can’t divide by 0. The function is discontinuous at $$x=0$$, with a non-removable discontinuity (but not an asymptote) at $$x=0$$, since the $$\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)=0$$ doesn’t exist. $$\displaystyle f\left( t \right)=\frac{t}{{2{{t}^{2}}+1}}$$ Since the denominator can never be 0 for real numbers (since of something squared added to something else), this function is continuous for all real $$t$$. $$f\left( x \right)=\left\{ \begin{array}{l}-3x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\le 3\\{{x}^{2}}+4x+1,\,\,\,\,x>3\end{array} \right.$$ Let’s check the “boundary point” at $$x=3$$; we see that for $$x=3$$, we have $$f\left( x \right)=-3\left( 3 \right)=-9$$. But the function “jumps” to the right of this point (for $$x>3$$), since we can see that $$f\left( x \right)={{\left( {3.01} \right)}^{2}}+4\left( {3.01} \right)+1=22.1001$$. The function is discontinuous (a non-removable discontinuity) at $$x=3$$. $$f\left( x \right)=\tan \left( x \right)$$ Since there are asymptotes at $$\displaystyle \frac{\pi }{2}+\pi k$$ for the tan function, the function is discontinuous at these values.

You might also see a problem like this:

 Function Solution If $$f\left( x \right)$$ is continuous at $$x=0$$, then what is the value of $$n$$:   \displaystyle f\left( x \right)=\left\{ \begin{align}&f\left( x \right)=\frac{{2{{x}^{3}}-x}}{{3x}},\,\,x\ne 0\\&f\left( 0 \right)=n\end{align} \right. We can’t just plug in 0 in the function, so let’s try to factor (take out the removable discontinuity): $$\require {cancel} \displaystyle \frac{{2{{x}^{3}}-x}}{{3x}}=\frac{{\cancel{x}\left( {2{{x}^{2}}-1} \right)}}{{3\cancel{x}}}=\frac{{2{{x}^{2}}-1}}{3}$$ Aha! Now, when $$x=0$$, we get $$\displaystyle f\left( 0 \right)=\frac{{2{{{\left( 0 \right)}}^{2}}-1}}{3}=-\frac{1}{3}$$. To make the function continuous, $$\displaystyle n=-\frac{1}{3}$$. You could also graph the function to get this result.

# Property of Limits

Limits have properties that are pretty straightforward; basically, these just mean that you can add, subtract, multiply, and divide limits (and multiply them by a number, or scalar) with the limit on the “inside” or “outside”. (Think about “picking apart” limits into smaller pieces.) And remember again that the limits refer to the “$$y$$” variable.

The properties are:

1. $$\underset{{x\to c}}{\mathop{{\lim }}}\,\left[ {b\cdot f\left( x \right)} \right]=b\cdot \underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)$$      (Scalar Multiple)
2. $$\underset{{x\to c}}{\mathop{{\lim }}}\left[ {f\left( x \right)\pm g\left( x \right)} \right]=\underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)\pm \underset{{x\to c}}{\mathop{{\lim }}}\,g\left( x \right)$$       (Sum or Difference)
3. $$\underset{{x\to c}}{\mathop{{\lim }}}\left[ {f\left( x \right)\cdot g\left( x \right)} \right]=\underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)\cdot \underset{{x\to c}}{\mathop{{\lim }}}\,g\left( x \right)$$       (Product)
4. $$\underset{{x\to c}}{\mathop{{\lim }}}\frac{{f\left( x \right)}}{{g\left( x \right)}}=\frac{{\underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)}}{{\underset{{x\to c}}{\mathop{{\lim }}}g\left( x \right)}}$$       (Quotient)
5. $$\underset{{x\to c}}{\mathop{{\lim }}}\,\left[ {f{{{\left( x \right)}}^{n}}} \right]={{\left[ {\underset{{x\to c}}{\mathop{{\lim }}}f\left( x \right)} \right]}^{n}}$$       (Power)
6. $$\underset{{x\to c}}{\mathop{{\lim }}}\,f\left( {g\left( x \right)} \right)=f\left( {\underset{{x\to c}}{\mathop{{\lim }}}g\left( x \right)} \right)$$        (Composite Functions)

Here is an example of how the sum property of limits works: $$\underset{{x\to 1}}{\mathop{{\lim }}}\left( {5{{x}^{2}}+2x-1} \right)=\underset{{x\to 1}}{\mathop{{\lim }}}5{{x}^{2}}+\underset{{x\to 1}}{\mathop{{\lim }}}2x-\underset{{x\to 1}}{\mathop{{\lim }}}1=5+2-1=6$$.

# Limits with Sine and Cosine

There are a couple of special trigonometric limits that you’ll need to know, and to use these, you may have to do some algebraic tricks. These are the two limits to learn:

$$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin x}}{x}=1\,\,\,\,\,\,\,\,\,\,\,\,(\text{thus}\underset{{x\to 0}}{\mathop{{\lim }}}\frac{x}{{\sin x}}=1)$$

$$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{1-\cos x}}{x}=0$$

Note that for the first limit (with sin), the reciprocal is also true, since  $$\displaystyle \frac{1}{1}=1$$.

Here are the types of problems you might see. Note again that you can check these in your calculator by putting in numbers really close to the $$x$$ values in your calculator (such as $$x=.00001$$ for x approaches 0).

 Trig Limit Solution $$\displaystyle \underset{{x\to \frac{\pi }{2}}}{\mathop{{\lim }}}\frac{{\sin x}}{x}$$ This problem looks like it may a case of the special trigonometric limits with sin, but since $$x$$ approaches $$\displaystyle \frac{\pi }{2}$$ instead of 0, we simply plug in $$\displaystyle \frac{\pi }{2}$$ in the top and bottom: $$\displaystyle \underset{{x\to \frac{\pi }{2}}}{\mathop{{\lim }}}\frac{{\sin x}}{x}=\frac{{\sin \frac{\pi }{2}}}{{\left( {\frac{\pi }{2}} \right)}}=\frac{1}{{\left( {\frac{\pi }{2}} \right)}}=\frac{2}{\pi }\approx .6366$$ Watch out for these sneaky problems! $$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{3x}}$$ We can use the special trigonometric limit in this problem, but we need to get it in the form $$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin x}}{x}$$. We can do this by using the Scalar Multiple Limit Property where we can use a cool trick:  move a $$\displaystyle \frac{1}{3}$$ to the outside, and then multiply by $$\displaystyle \frac{4}{4}$$ (which is 1) to get a $$4x$$ on the bottom: \displaystyle \begin{align}\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{3x}}&=\frac{1}{3}\cdot \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{x}=\frac{1}{3}\cdot \frac{4}{4}\cdot \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{x}\\&=\frac{1}{3}\cdot \frac{4}{1}\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{4x}}=\frac{4}{3}\cdot 1=\frac{4}{3}\end{align} Notice in these problems that we can sort of look at the fraction in the problem ($$\displaystyle \frac{4}{3}$$) and that will be the answer, but you’ll probably have to show your work! $$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{{{{\left( {1-\cos x} \right)}}^{2}}}}{{2x}}$$ We can see how this might be a case of the special trigonometric limits with cos, so let’s try it and see how to use some algebra to figure it out. Notice how we use the Product Limit Property to separate the expression: \displaystyle \begin{align}\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{{{{\left( {1-\cos x} \right)}}^{2}}}}{{2x}}&=\underset{{x\to 0}}{\mathop{{\lim }}}\left( {\frac{{1-\cos x}}{2}} \right)\left( {\frac{{1-\cos x}}{x}} \right)\\&=\underset{{x\to 0}}{\mathop{{\lim }}}\left( {\frac{{1-\cos x}}{2}} \right)\cdot \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\frac{{1-\cos x}}{x}} \right)\\&=\left( {\frac{{1-\cos 0}}{2}} \right)\cdot 0=\frac{0}{2}\cdot 0=0\end{align} $$\displaystyle \underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\tan t}}{{2t}}$$ This one’s a little tricky since we don’t have a special trig limit for tan per se, and we will end up with a 0 in the denominator.  Since $$\displaystyle \tan =\frac{{\sin }}{{\cos }}$$,  let’s turn tan into sin and cos: \begin{align}\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\tan t}}{{2t}}&=\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\frac{{\sin t}}{{\cos t}}}}{{2t}}=\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\sin t}}{{\cos t\cdot 2t}}\\&=\underset{{t\to 0}}{\mathop{{\lim }}}\left( {\frac{{\sin t}}{{2t}}\cdot \frac{1}{{\cos t}}} \right)=\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\sin t}}{{2t}}\cdot \underset{{t\to 0}}{\mathop{{\lim }}}\frac{1}{{\cos t}}\\&=\left( {\frac{1}{2}} \right)\underset{{t\to 0}}{\mathop{{\lim }}}\frac{{\sin t}}{t}\cdot \underset{{t\to 0}}{\mathop{{\lim }}}\frac{1}{{\cos t}}=\frac{1}{2}\cdot 1=\frac{1}{2}\end{align} So if we see a trig function other than sin or cos, sometimes we can turn it into something that works! $$\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sin \left( {3h} \right)}}{{\sin \left( {2h} \right)}}$$ Since $$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin x}}{x}=\underset{{x\to 0}}{\mathop{{\lim }}}\frac{x}{{\sin x}}=1$$, we can turn this limit into two limits with matching sin arguments and denominators, and separate the limits: \begin{align}\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sin \left( {3h} \right)}}{{\sin \left( {2h} \right)}}&=\underset{{h\to 0}}{\mathop{{\lim }}}\left( {\frac{3}{2}} \right)\left( {\frac{{2h}}{{3h}}} \right)\frac{{\sin \left( {3h} \right)}}{{\sin \left( {2h} \right)}}=\frac{3}{2}\underset{{h\to 0}}{\mathop{{\lim }}}\left( {\frac{{\sin \left( {3h} \right)}}{{3h}}\cdot \frac{{2h}}{{\sin \left( {2h} \right)}}} \right)\\&=\frac{3}{2}\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sin \left( {3h} \right)}}{{3h}}\cdot \underset{{h\to 0}}{\mathop{{\lim }}}\frac{{2h}}{{\sin \left( {2h} \right)}}=\frac{3}{2}\cdot 1\cdot 1=\frac{3}{2}\end{align} $$\underset{{t\to 0}}{\mathop{{\lim }}}t\sec t$$ Let’s turn the sec into cos to see if we can find a special trig limit. As it turns out, we don’t have to use a special limit, since we can plug $$t=0$$ into both the numerator and denominator: $$\displaystyle \underset{{t\to 0}}{\mathop{{\lim }}}\left( {t\sec t} \right)=\underset{{t\to 0}}{\mathop{{\lim }}}\frac{t}{{\cos t}}=\frac{0}{{\cos \left( 0 \right)}}=\frac{0}{1}=0$$

# Intermediate Value Theorem (IVT)

The intermediate value theorem (IVT) seems very complicated and is a bit theoretical, but if we think about what it really says, it’s not that difficult and pretty obvious.

What the intermediate value theorem says is that if you are at a certain $$x$$ (where you have a $$y$$ value)and you go to another value $$x$$ to the right (where you have another $$y$$ value), and the path that you go is on a continuous function, then you have to have hit (cross over) all the $$y$$ values in between.

So that still sounds confusing, so let’s think of an example. Let’s say a baby boy weighs 7 pounds at birth, and then 20 pounds when he is 1 year old (12 months). At some point, he must have weighed say 15 pounds, or actually any number of pounds between 7 and 20 pounds. In this case, the baby’s age is the $$x$$ value, and the baby’s weight is the $$y$$ value, with the interval being between 0 and 12 months, inclusive. This makes sense, since a human weight is continuous, and it doesn’t jump up or down instantaneously.

The other way to think of IVT is that we have 2 points on a continuous curve and there is a horizontal line between these two points, then the curve must cross this horizontal line to get from one point to the other point.

Here is the formal definition (and picture) of the Intermediate Value Theorem:

 Intermediate Value Theorem:   If a function $$f$$ is continuous on a closed interval $$[a,b]$$, where $$f\left( x \right)={{x}^{2}}-2x-3$$, and $$m$$ is any number between $$f\left( a \right)$$ and $$f\left( b \right)$$, there must be at least one number $$c$$ in $$[a,b]$$ such that $$f\left( c \right)=m$$.

Here are some types of problems that you might see with the Intermediate Value Theorem:

IVT Problem Solution
Explain why the following function has a zero in the given interval:

$$f\left( x \right)={{x}^{2}}-2x-3$$ in interval $$[2,5]$$

Since $$f\left( x \right)={{x}^{2}}-2x-3$$ is continuous on the interval $$[2,5]$$, and $$f\left( 2 \right)=-3$$, and $$f\left( 5 \right)=12$$, by the Intermediate Value Theorem (IVT), there exists a number $$c$$ in $$[2,5]$$ such that $$f\left( c \right)=0$$ (0 is between –3 and 12).
Explain why the following function has a zero in the given interval:

$$f\left( x \right)={{x}^{2}}-1-\sin \left( x \right)$$ in interval $$\left[ {-\pi ,0} \right]$$

$$f\left( x \right)={{x}^{2}}-1-\sin \left( x \right)$$ is continuous on the interval $$\left[ {-\pi ,0} \right]$$ (you can check on a graphing calculator).

$$f\left( {-\pi } \right)={{\left( {-\pi } \right)}^{2}}-1-\sin \left( {-\pi } \right)={{\pi }^{2}}-1-0\approx 8.9$$, and $$f\left( 0 \right)={{\left( 0 \right)}^{2}}-1-\sin \left( 0 \right)=0-1-0=-1$$, so by the IVT, there exists a number $$c$$ in $$\left[ {-\pi ,0} \right]$$ such that $$f\left( c \right)=0$$ (0 is between –1 and 8.9).

Does the Intermediate Value Theorem guarantee a “$$c$$” such that $$f\left( c \right)=0$$ for $$f\left( x \right)={{x}^{3}}-{{x}^{2}}+2x-2$$ on $$[–1,5]$$?

If so, explain how you know, and find that “$$c$$” value. If not, explain why.

We notice that the function is continuous on $$[–1,5]$$, so we can go on. We plug our endpoints into the function to see what the values of $$f\left( x \right)$$ is at those points: $$f\left( {-1} \right)={{\left( {-1} \right)}^{3}}-{{\left( {-1} \right)}^{2}}+2\left( {-1} \right)-2=-6$$ and $$f\left( 5 \right)={{\left( 5 \right)}^{3}}-{{\left( 5 \right)}^{2}}+2\left( 5 \right)-2=108$$. Since –6 is negative and 108 is positive, by IVT, there exists a number $$c$$ in $$[–1,5]$$ such that $$f\left( c \right)=0$$.

To find that “$$c$$”, simply set the original function to 0 and solve for $$x$$:

$$\begin{array}{c}{{x}^{3}}-{{x}^{2}}+2x-2=0\\{{x}^{2}}\left( {x-1} \right)+2\left( {x-1} \right)=0\\\left( {{{x}^{2}}+2} \right)\left( {x-1} \right)=0\\x=1\end{array}$$

We used grouping to  factor the polynomial, and set each factor to 0. We see that $$x=1$$ is the only solution to the function, and since $$-6<1<108$$, $$c=1$$.

Does the Intermediate Value Theorem guarantee a “$$c$$” such that $$f\left( c \right)=2$$ for $$f\left( x \right)={{x}^{3}}-3x-2$$ on $$[–3,0]$$?

If so, explain how you know, and find that “$$c$$” value.  If not, explain why.

We notice that the function is continuous on $$[–3,0]$$, so we can go on. We plug in our endpoints into the function to see what the values of $$f\left( x \right)$$ will be at those points: $$f\left( {-3} \right)={{\left( {-3} \right)}^{3}}-\left( {-3} \right)-2=-26$$ and $$f\left( 0 \right)={{\left( 0 \right)}^{2}}-\left( 0 \right)-2=-2$$.

Since $$f\left( c \right)=2$$ doesn’t fall into the interval –26 to –2, there is not necessarily a number $$c$$ in $$[–3,0]$$  such that $$f\left( c \right)=2$$. We can stop here.

If $$f\left( x \right)$$ and $$g\left( x \right)$$ are continuous functions with the values below.

Does $$h\left( x \right)$$ have a zero on the interval $$[2,4]$$, given $$h\left( x \right)=f\left( {g\left( x \right)} \right)$$? Why or why not?

 $$x$$ $$f\left( x \right)$$ $$g\left( x \right)$$ 1 7 3 2 6 3 3 –2 1 4 3 1
To find out if $$h\left( x \right)$$ has a zero on the interval $$[2,4]$$, let’s put in the endpoints and see what those values are. $$h\left( 2 \right)=f\left( {g\left( 2 \right)} \right)=f\left( 3 \right)=-2$$ (remember to work from the inside out), and $$h\left( 4 \right)=f\left( {g\left( 4 \right)} \right)=f\left( 1 \right)=7$$.

Since $$f\left( x \right)$$ and $$g\left( x \right)$$ are both continuous, then $$h\left( x \right)=f\left( {g\left( x \right)} \right)$$ is continuous. Then by the IVT, there exists a zero on the interval $$[2,4]$$, since $$-2<0<7$$ (it must cross the $$x$$-axis).

# Infinite Limits

An infinite limit is just a limit in which the $$y$$ either increases or decreases without bound (goes up forever or down forever) as $$x$$ gets closer and closer to a value. We typically think of these types of limits when we deal with vertical asymptotes (VA’s), so we can use what we know about VA’s to work with them. As an example, these limits exist in rational functions with a denominator of 0.

When a function gets closer and closer to a VA from one side, or both sides (if the limit exists), the limit will either be $$-\infty$$ or $$\infty$$. (Theoretically, the limit doesn’t exist since these aren’t real numbers, but we still say these limits are $$-\infty$$ or $$\infty$$). To determine which one it is, we can put in numbers (for $$x$$) really close to the VA on either side (to see what direction the graph is going), or use a graphing calculator. Sometimes, we can use algebra to simplify a rational function to get the limit.

Let’s do some problems where we need to find the one-sided limit (if it exists). Some of these may involve remembering rational parent functions. You can also try these on your graphing calculator to get the answers.

 Limit Problem Solution Find the one-sided limit:   $$\displaystyle \underset{{x\to -{{2}^{+}}}}{\mathop{{\lim }}}\frac{1}{{x+2}}$$ Shifting the graph 2 units to the left from the parent function $$\displaystyle \frac{1}{x}$$, we get: Coming in from the right (because of the small $$+$$ sign) to –2 (vertical asymptote), notice that the $$y$$ is getting more and more positive (closer and closer to $$\infty$$) as it gets closer to –2. Therefore, $$\displaystyle \underset{{x\to -{{2}^{+}}}}{\mathop{{\lim }}}\frac{1}{{x+2}}=\infty$$. (We also could have also plugged in values for $$x$$ close to –2  from the positive side, like –1.8 and –1.9, and notice the $$y$$ is increasing). Note that (the two-sided) $$\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\frac{1}{{x+2}}$$ does not exist, since the limit from the left is $$-\infty$$, and from the right is $$+\infty$$. Find the limit:   $$\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{1}{{{{{\left( {x+2} \right)}}^{2}}}}$$ This isn’t a one-sided limit, but it still exists! Because of the square in the denominator, the function hugs the vertical asymptote in the same direction; thus, $$\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{1}{{{{{\left( {x+2} \right)}}^{2}}}}=\infty$$. Find the one-sided limit:   $$\displaystyle \underset{{x\to \,{{3}^{-}}}}{\mathop{{\lim }}}\,\frac{x}{{3-x}}$$ We see that the vertical asymptote is $$x=3$$, and to find out where the graph is coming from on the left side of 3 (because of the minus), let’s plug in 2.8 and 2.9: we notice the $$y$$ is getting larger. Thus, the value of $$y$$ will get more and more positive as it comes from the left and gets closer to 3. Therefore, $$\displaystyle \underset{{x\to {{3}^{-}}}}{\mathop{{\lim }}}\frac{x}{{3-x}}=\infty$$. Find the one-sided limit:   $$\displaystyle \underset{{x\to \,{{3}^{-}}}}{\mathop{{\lim }}}\,\frac{{x-3}}{{{{x}^{2}}-x-6}}$$ Factor and simplify; we see we have a removable discontinuity, or hole, at 3, and a vertical asymptote at –2: $$\require{cancel} \displaystyle \frac{{x-3}}{{{{x}^{2}}-x-6}}=\frac{{\cancel{{x-3}}}}{{\cancel{{\left( {x-3} \right)}}\left( {x+2} \right)}}=\frac{1}{{x+2}}$$ Now we can plug in $$3$$ for $$x$$ (even though the limit comes from the left), to get $$\displaystyle \frac{1}{{3+2}}=\frac{1}{5}$$. Therefore, $$\displaystyle \underset{{x\to {{3}^{-}}}}{\mathop{{\lim }}}\frac{{x-3}}{{{{x}^{2}}-x-6}}=\frac{1}{5}$$. Find the one-sided limit:   $$\displaystyle \underset{{x\to \,{{1}^{+}}}}{\mathop{{\lim }}}\frac{{{{x}^{2}}-3}}{{x-1}}$$ Let’s just graph this one; we see that $$\displaystyle \underset{{x\to \,{{1}^{+}}}}{\mathop{{\lim }}}\frac{{{{x}^{2}}-3}}{{x-1}}=-\infty$$. Find the one-sided limit:   $$\displaystyle \underset{{x\to \,{{1}^{+}}}}{\mathop{{\lim }}}\,3\ln \left( {x-1} \right)+2$$ We know that the vertical asymptote in the parent function $$y=\ln x$$ is $$x=0$$, but this function is shifted to the right 1. Therefore, $$\displaystyle \underset{{x\to \,{{1}^{+}}}}{\mathop{{\lim }}}\,3\ln \left( {x-1} \right)+2=-\infty$$. Find the one-sided limit:   $$\displaystyle \underset{{x\to \,{{{\frac{\pi }{2}}}^{+}}}}{\mathop{{\lim }}}\,\left( {\sec \left( x \right)} \right)$$ If we plug in $$\displaystyle \frac{\pi }{2}$$ for $$x$$ in $$\sec \left( x \right)$$, it is undefined, as we have an asymptote at $$\displaystyle x=\frac{\pi }{2}$$. But notice as $$x$$ comes from the right (the small $$+$$ sign), the $$y$$ is getting more and more negative as the graph gets closer to $$\displaystyle \frac{\pi }{2}$$. (We could also plug in a number a little larger than $$\displaystyle \frac{\pi }{2}$$ to see what value we get). Therefore, $$\displaystyle \underset{{x\to {{{\frac{\pi }{2}}}^{+}}}}{\mathop{{\lim }}}\left( {\sec \left( x \right)} \right)=-\infty$$.

# Limits at Infinity

Limits at Infinity exist when the $$x$$ values (not the $$y$$) go to $$\infty$$ or $$-\infty$$. This can happen when we work with rational functions and we have more one or more with horizontal asymptotes (HAs) (which are end behavior asymptotes, or EBAs). The $$y$$ values can get closer and closer to a number, but never actually reach that number in the case of an EBA. Let’s review how to get horizontal or end behavior asymptotes:

 Rules for Graphing Rationals Example To get the end behavior asymptote (EBA), you want to compare the degree in the numerator to the degree in the denominator. There can be at most 1 EBA and, most of the time, these are horizontal.   ⇒  If the degree (largest exponent) on the bottom is greater than the degree on the top, the EBA (which is also a horizontal asymptote or HA) is $$y=0$$. $$\displaystyle y=\frac{{x+2}}{{{{x}^{2}}-4}}$$   Notice that even though we can take out a removable discontinuity ($$x+2$$), the bottom still has a higher degree than the top, so the HA/EBA is $$y=0$$. ⇒  If the degree on the top is greater than the degree on the bottom, there is no HA. However, if the degree on the top is one more than the degree on the bottom, there is a slant (oblique) EBA asymptote, which is discussed below. $$\displaystyle y=\frac{{{{x}^{3}}+2}}{{x-4}}$$   No HA/EBA. Vertical asymptote is still $$\displaystyle x=4$$. ⇒  If the degree is the same on the top and the bottom, divide coefficients of the variables with the highest degree on the top and bottom; this is the HA/EBA. You can determine this asymptote even without factoring. $$\displaystyle y=\frac{{2{{x}^{3}}+2}}{{3{{x}^{3}}-4}}$$   Since the degree on the top and bottom are both 3, the HA/EBA is $$\displaystyle y=\frac{2}{3}$$. ⇒  If the degree on the top is one more than the degree on the bottom, then the function has a slant or oblique EBA in the form $$y=mx+b$$. We have to use long division to find this linear equation.   We can just ignore or “throw away” the remainder and just use the linear equation. Weird, huh? $$\displaystyle y=\frac{{2{{x}^{2}}+x+1}}{{x-4}}$$         $$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,+9\\x-4\overline{\left){{2{{x}^{2}}+x+1}}\right.}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{2{{x}^{2}}-8x\,\,\,\,\,\,}}\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x+1\end{array}$$                 EBA:   $$y=2x+9$$

The easiest way to get limits at infinity with rational functions is to find the horizontal asymptotes in that direction. We can also use a trick where we can divide every term in the numerator and denominator by the variable with the highest degree (highest exponent value). This is because of the following Limits of Infinity Theorems:

$$\displaystyle \underset{{x\to \,\infty }}{\mathop{{\lim }}}\,\frac{a}{{{{x}^{r}}}}=0;\,\,\,\,\,\underset{{x\to \,-\infty }}{\mathop{{\lim }}}\,\frac{a}{{{{x}^{r}}}}=0$$,     $$a$$ is a real number, $$r$$ is a positive rational number

Basically all this says is that if the bottom (numerator) of a fraction gets bigger and bigger (towards $$\infty$$ or $$-\infty$$), the whole fraction will get smaller and smaller and eventually go to 0.

When we do limit problems where there are $$x$$’s on the top and bottom, when we try to plug in $$\infty$$ or $$-\infty$$, we’ll typically get what we call indeterminate form – something like  $$\displaystyle \frac{\infty }{\infty }$$. So we’ll have to use the tricks of finding the horizontal asymptote, or dividing all the terms by the variable with the highest degree. Sometimes if we have roots in the function, we can multiply by the conjugate of the numerator or denominator and try to go from there.

Note that you can check these by trying to put in a large number (or very small number for $$-\infty$$) for $$x$$ in your graphing calculator.

Also note that with Limits at Infinity, if there is no horizontal asymptote (the degree on the bottom is less than the degree on the top with a rational function), the limit doesn’t exist.

Let’s do some problems.

 Limit Problem Solution Find the limit:   $$\displaystyle \underset{{x\to \,\infty }}{\mathop{{\lim }}}\,\frac{{3x-1}}{{x+2}}$$ We can see that we have a horizontal EBA (end behavior asymptote) at $$y=3$$, since when the exponents are the same on the top and bottom, we divide coefficients. We can also divide each term by $$x$$, since this is the highest degree (exponent): $$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{{3x-1}}{{x+2}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{{3x}}{x}-\frac{1}{x}}}{{\frac{x}{x}+\frac{2}{x}}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\frac{{3-\frac{1}{x}}}{{1+\frac{2}{x}}}=\frac{{3-0}}{{1+0}}=3$$. So $$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{{3x-1}}{{x+2}}=3$$. Find the limit:   $$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{x}{{{{x}^{2}}-1}}$$ We can see that we have a horizontal asymptote at $$y=0$$, since the degree is larger on the bottom. We can also divide each term by $$x$$, since this is the highest degree (exponent): $$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{x}{{{{x}^{2}}-1}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\frac{{\frac{x}{{{{x}^{2}}}}}}{{\frac{{{{x}^{2}}}}{{{{x}^{2}}}}-\frac{1}{{{{x}^{2}}}}}}=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{x}}}{{1-\frac{1}{{{{x}^{2}}}}}}=\frac{0}{{1-0}}=0$$ Find the limit:   $$\displaystyle \underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{{2x-1}}{{\sqrt{{{{x}^{2}}+x-3}}}}$$ The degree on top and bottom are the same, so if we divide coefficients of $$\displaystyle \frac{{2x}}{{\sqrt{{{{x}^{2}}}}}}$$ to get the EBA, we get 2. We need to make this negative since the limit is going to $$-\infty$$. so we get –2 ($$\displaystyle \sqrt{{{{x}^{2}}}}$$ is always positive). We could also write $$x$$ as $$\displaystyle \sqrt{{{{x}^{2}}}}$$ (actually, we have to use $$\displaystyle -\sqrt{{{{x}^{2}}}}$$, since the limit is going to $$-\infty$$, so $$\displaystyle x<0$$), and divide each term by either $$x$$ or  (use $$\displaystyle -\sqrt{{{{x}^{2}}}}$$ for terms under the square roots): \displaystyle \begin{align}\underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{{\frac{{2x}}{x}-\frac{1}{x}}}{{\frac{{\sqrt{{{{x}^{2}}+x-3}}}}{{-\sqrt{{{{x}^{2}}}}}}}}&=\underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{{2-\frac{1}{x}}}{{-\sqrt{{\frac{{{{x}^{2}}+x-3}}{{{{x}^{2}}}}}}}}=\underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{{2-\frac{1}{x}}}{{-\sqrt{{1+\frac{1}{x}-\frac{3}{{{{x}^{2}}}}}}}}\\&=\frac{{2-0}}{{-\sqrt{{1+0-0}}}}=-2\end{align} Find the limit:   $$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\left( {x-\sqrt{{{{x}^{2}}+x}}} \right)$$ We can plug in huge numbers for $$x$$ and check this on the graphing calculator, but using algebra, we have to do something tricky: treat the function as a fraction (over 1), and rationalize the numerator (multiply the numerator and denominator by the conjugate of the numerator). Again, we’ll want to write $$x$$ as $$\displaystyle \sqrt{{{{x}^{2}}}}$$, which is positive, because of $$x$$ going to $$\infty$$: \displaystyle \begin{align}\underset{{x\to \infty }}{\mathop{{\lim }}}\left( {x-\sqrt{{{{x}^{2}}+x}}} \right)&=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{x-\sqrt{{{{x}^{2}}+x}}}}{1}} \right)\left( {\frac{{x+\sqrt{{{{x}^{2}}+x}}}}{{x+\sqrt{{{{x}^{2}}+x}}}}} \right)\\&=\underset{{x\to \infty }}{\mathop{{\lim }}}\left( {\frac{{{{x}^{2}}-{{x}^{2}}-x}}{{x+\sqrt{{{{x}^{2}}+x}}}}} \right)=\underset{{x\to \infty }}{\mathop{{\lim }}}\left( {\frac{{-x}}{{x+\sqrt{{{{x}^{2}}+x}}}}} \right)\\&=\underset{{x\to \infty }}{\mathop{{\lim }}}\left( {\frac{{-\frac{x}{x}}}{{\frac{x}{x}+\frac{{\sqrt{{{{x}^{2}}+x}}}}{{\sqrt{{{{x}^{2}}}}}}}}} \right)=\underset{{x\to \infty }}{\mathop{{\lim }}}\left( {\frac{{-1}}{{1+\sqrt{{\frac{{{{x}^{2}}+x}}{{{{x}^{2}}}}}}}}} \right)\\&=\underset{{x\to \infty }}{\mathop{{\lim }}}\left( {\frac{{-1}}{{1+\sqrt{{1+\frac{1}{x}}}}}} \right)=-\frac{1}{{1+\sqrt{{1+0}}}}=-\frac{1}{2}\end{align} Find the limit:   $$\displaystyle \underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{{4{{x}^{2}}}}{{x+1}}$$ We can see from the asymptote rules that we don’t have a horizontal asymptote (we have an oblique asymptote for the EBA). Let’s still divide all terms by the highest degree: $$\displaystyle \underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{{4{{x}^{2}}}}{{x+1}}=\underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{{\frac{{4{{x}^{2}}}}{{{{x}^{2}}}}}}{{\frac{{x+1}}{{{{x}^{2}}}}}}=\underset{{x\to -\infty }}{\mathop{{\lim }}}\frac{4}{{\frac{1}{x}+\frac{1}{{{{x}^{2}}}}}}=\frac{4}{{0+0}}=-\infty$$ We get $$-\infty$$, by looking at a graph. Nonetheless no limit exists! Find the limit:   $$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{1}{{3x+\sin x}}$$ We can see that the denominator will get bigger and bigger as $$x$$ goes towards $$\infty$$, since $$-1\le \sin x\le 1$$. So $$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{1}{{3x+\sin x}}=0$$. Try this on your graphing calculator!

# Limits of Sequences

We looked at Sequence and Series here, but since a sequence is really a function using the natural numbers as $$x$$. So we can think of the limit of a sequence (if that limit exists) as what $$y$$-value is approaching, as the number of terms goes to $$\infty$$ (similar to Limits at Infinity above). This is the number to which this sequence converges.

We learned in the Sequences and Series section that all arithmetic sequences diverge, and geometric sequences converge if $$\left| r \right|<1$$, where $$r$$ is the common ratio; actually, these types of sequences converge to 0. For other sequences, we can use the same rules that we used for Limits at Infinity above, since those limits were also going to $$\infty$$. We are really just looking for the end behavior of the sequences; what our $$y$$ is converging to, when our $$x$$ gets bigger and bigger.

For example, the limit of the sequence $$\displaystyle {{a}_{n}}=\frac{1}{{{{n}^{2}}}}\,\,\,(1,\frac{1}{4},\frac{1}{9},\frac{1}{{16}},…)$$ ($$n$$ is a positive integer) is 0, since the $$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{1}{{{{n}^{2}}}}=0$$. Therefore, this sequence converges to 0.

Here are more examples of limits of sequences:

 Sequence Limit Find the limit: $$\displaystyle {{a}_{n}}=4{{\left( {.6} \right)}^{{n-1}}}$$ This is a geometric sequence in the form $${{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n-1}}}$$. Since $$\left| {.6} \right|<1$$, this sequence converges to 0. Note: To see what the series ($${{S}_{\infty }}=\sum\limits_{{k=1}}^{\infty }{{4{{{\left( {.6} \right)}}^{{n-1}}}}}$$) would converge to, we would use the equation $$\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1-r}}=\frac{4}{{1-.6}}=10$$. Find the limit: $$\displaystyle {{a}_{n}}=.5{{\left( {\frac{9}{8}} \right)}^{{n-1}}}$$ This is a geometric sequence in the form $${{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n-1}}}$$. Since $$\displaystyle \left| {\frac{9}{8}} \right|>1$$, this sequence diverges. Find the limit: $${{a}_{n}}=4n+1$$ This is an arithmetic sequence, so it diverges. Find the limit: $$\displaystyle {{a}_{n}}=\frac{{6{{n}^{2}}+3}}{{n-5}}$$ This is neither arithmetic nor geometric, but the degree on the top is greater than the degree on the bottom; based on the rules above, no limit exists. Therefore, the sequence diverges. Find the limit: $$\displaystyle {{a}_{n}}=\frac{{6n-4}}{{5n+3}}$$ This is neither arithmetic nor geometric, but the degree on the top is the same as the degree on the bottom; based on the rules above, divide coefficients of the variables with the highest degree on the top and bottom; this is the HA/EBA. The sequence converges to its EBA, $$\displaystyle \frac{6}{5}$$. Find the limit: $$\displaystyle {{a}_{n}}=\frac{{8{{n}^{2}}+7}}{{{{n}^{3}}-3}}$$ This is neither arithmetic nor geometric, but the degree on the top is less than the degree on the bottom; based on the rules above, the limit is 0. Therefore, the sequence converges to 0. Find the limit: $$\displaystyle {{a}_{n}}=-1,2,-1,…..$$ The sequence just goes back and forth between two numbers, so it never converges. Therefore, the sequence diverges. Find the limit: $$\displaystyle {{a}_{n}}=1,3,5,7,…..$$ This is an arithmetic sequence, so it diverges. Find the limit: $$\displaystyle {{a}_{n}}=-1,2,-\frac{1}{2},\frac{3}{2},-\frac{1}{4},\frac{3}{4}…$$ Even though the sequence oscillates between negative and positive numbers, it still gets closer and closer to 0, so the sequences converges to 0. Find the limits: $$\displaystyle \begin{array}{l}{{a}_{n}}=\cos \left( n \right)\\{{a}_{n}}=\sin \left( n \right)\end{array}$$ These sequences are periodic and oscillate, so they never converge (the sequences diverge).

Learn these rules, and practice, practice, practice!

Use the MathType keyboard to enter a Limit problem, and then click on Submit (the arrow to the right of the problem) to solve the problem. You can also click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device.  Enjoy!

On to Definition of the Derivative  – you are ready!

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