Note that we discuss finding limits using L’Hopital’s Rule here. Also note that there’s a very good limit calculator here on the this online calculator site.
Introduction to Limits
We need to understand how limits work, since the first part of Differential Calculus uses them extensively. (Differential calculus has to do with rates at which quantities change.) 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). Again, remember that limits are always the $ y$-value (dependent variable), not the $ x$-value (independent variable).
We can write a limit where $ x$ gets closer and closer to 0 as $ \lim\limits_{x \to 0}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 section). Even if a function is “normal”, like a linear function, we still consider the $ y$-value a limit where it touches the $ x$-value, as shown below.
The reason we have limits in Differential Calculus is because sometimes we need to know what happens to a function when $ 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 would typically be undefined.
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 from the Advanced Factoring section in this example.
| Limit | Graph | ||||||||||||||||
| $ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\frac{{{{x}^{3}}-1}}{{x-1}}=3$
T-Chart:
| 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! Thus, $ \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 problems. For limits with trigonometry, see below.
| Rule | Problem | Solution |
| Simply Plugging in $ x$-value | $ \lim\limits_{x \to 4}\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; factor the top (difference of cubes), simplify, and then plug in 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, 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’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, 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.
The limit from the right, or $ \lim\limits_{x \to {{{c}^{+}}}}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 $ \lim\limits_{x \to {{{c}^{-}}}}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
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 concept leads to the definition of the existence of a limit, the formal definition of continuity:
| Limit Theorem/Definition | Explanation |
| Existence of a Limit
$ \lim\limits_{x \to c}f\left( x \right)=L$ if and only if
$ \lim\limits_{x \to {{c}^{-}}}f\left( x \right)=L$ and $ \lim\limits_{x \to {{c}^{+}}}f\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 $ f\left( x \right)$ at that point. The actual point $ f\left( c \right)$ may be defined (as in a non-continuous function) at a completely different $ y$-value, where no limit may occur.
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 )$; in this case, there would be an infinite limit. |
| Definition of Continuity
A function is continuous at $ c$ when the following three conditions are met:
| 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$-values. 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 (functions with 0 in the denominator); these functions are discontinuous at these points. Let’s review how we get vertical asymptotes for a rational function:
| Holes and Vertical Asymptotes | Examples |
| Check for removable discontinuities (holes) 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 (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. ⇒ (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 the reduced fraction). Therefore, 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. We have to “skip over” –3, 0, and 3. |
And we’ll also have to remember the Trigonometric Function asymptotes (remember, the ones that start with “C” are the easiest (“eaCest”):
| 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 (has a limit) or non-removable (doesn’t have a limit):
| Function | Continuous or Discontinuous? |
| $ \displaystyle f\left( x \right)=\frac{x}{{{{x}^{2}}+2x}}$ | Factor the denominator to see there is one removable discontinuity at $ x=0$, and one non-removable discontinuity (vertical asymptote) at $ 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)}}$ The function is discontinuous at $ x=0$ and $ x=-2$, where the denominator $ =0$. |
| $ \displaystyle f\left( x \right)=\frac{{\left| x \right|}}{x}$ | First turn this into a piecewise function, with the “boundary point” at $ x=0$ (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 the denominator is 0. Thus, the function is discontinuous at $ x=0$, with a non-removable discontinuity (but not an asymptote) at $ x=0$, since the $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)$ doesn’t exist. Graph the function to see this! |
| $ \displaystyle f\left( t \right)=\frac{t}{{2{{t}^{2}}+1}}$ | Since the denominator can never be 0 for real numbers (since $ {{t}^{2}}\ge 0$, making $ 2{{t}^{2}}+1$ positive), 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.$ | For the function to be continuous, it must have the same value at the “boundary point”, which is $ x=3$. For the first part of the function, $ f\left( 3 \right)=-3\left( 3 \right)=-9$, but for the second part, $ f\left( 3 \right)={{\left( 3 \right)}^{2}}+4\left( 3 \right)+1=22$. 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 has non-removable discontinuities 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 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}$. You could also graph the function to get this result. |
Property of Limits
Limits have properties that are pretty straightforward; basically, you can add, subtract, multiply, and divide limits (and multiply them by a number, or scalar) with the limit on the “inside” or “outside”. (Think about “picking apart” limits into smaller pieces.) Remember again that limits refer to the “$ y$”-value, or the dependent variable.
The properties are:
- $ \displaystyle \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)
- $ \displaystyle \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)
- $ \displaystyle \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)
- $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\frac{{f\left( x \right)}}{{g\left( x \right)}}=\frac{\lim\limits_{x \to c}f(x)}{{\lim\limits_{x \to c}g(x)}}$ (Quotient)
- $ \displaystyle \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)
- $ \displaystyle \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 Function)
Here is an example of how the sum property of limits works: $ \displaystyle \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$.
Squeeze Theorem
The Squeeze Theorem (or Sandwich Theorem) is used to find the limit of a function, say $ f\left( x \right)$, at a particular point by showing that if two functions “on the outside of” that function have a limit at a certain point, $ f\left( x \right)$ has that same limit. In other words, $ f\left( x \right)$ is “trapped” between two functions with a certain limit, so it must have that same limit. The concept is based on this: $ \text{if }a\le b\le c\text{,}\,\,\text{and}\,\,a=c,\,\,\text{then}\,a=b=c\,$. Thus:
If $ f\left( x \right)\le g\left( x \right)\le h\left( x \right)$, and $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f\left( x \right)=\underset{{x\to c}}{\mathop{{\lim }}}\,h\left( x \right)=L,\,\,\underset{{x\to c}}{\mathop{{\lim }}}\,g\left( x \right)=L$
Here is a visual representation of what the function $ g\left( x \right)=\left| x \right|\cos x$ might look like between the functions $ f\left( x \right)=\left| x \right|$ and $ h\left( x \right)=-\left| x \right|$, all three of which functions have a limit of $ 0$ as $ x\to 0$. It is solved algebraically below.
Here are some types of problems you may see:
| Squeeze Theorem Problem | Solution |
| Find $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\left| x \right|\cos x} \right)$ | The trick is to use something we know: that $ \cos x$ is between $ -1$ and $ 1$, and then manipulate it to turn into what we need. Then we can take the limit of all three functions so that the function we need is in the middle (thus, “squeezed” in): $ \begin{align}-1&\le \cos x\le 1\\-\left| x \right|&\le \left| x \right|\cos x\le \left| x \right|\\\underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {-\left| x \right|} \right)&\le \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\left| x \right|\cos x} \right)\le \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\left| x \right|} \right)\\0&\le \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\left| x \right|\cos x} \right)\le 0\\\text{Thus},\,&\underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\left| x \right|\cos x} \right)=0\,\end{align}$ |
| Use the squeeze theorem to find $ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f\left( x \right)$: $ \begin{array}{c}c=0\\3-{{x}^{2}}\le f\left( x \right)\le 3+{{x}^{2}}\end{array}$ | $ \begin{align}\underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {3-{{x}^{2}}} \right)&\le \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)\le \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {3+{{x}^{2}}} \right)\\3&\le \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)\le 3\\\text{Thus},\,&\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)=3\end{align}$ |
The squeeze theorem is helpful with functions that oscillate, and is used to prove the following limits with sine and cosine functions.
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\,\,\,\,\,\,\,\,\,\,(\underset{{x\to 0}}{\mathop{{\lim }}}\frac{x}{{\sin x}}=1)$
$ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{1-\cos x}}{x}=\underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\cos x-1}}{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 (for example, $ x=.00001$ for $ x$ approaching 0).
| Trig Limit Problem | 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 ($ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin \,x}}{x}=1$), 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}}$ | Use the special trigonometric limit in this problem, but first get it in the form $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{\sin x}}{x}$. Use the Scalar Multiple Limit Property and a 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}\left( {\underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{x}} \right)=\frac{1}{3}\cdot \frac{4}{4}\left( {\underset{{x\to \,0}}{\mathop{{\lim }}}\,\frac{{\sin 4x}}{x}} \right)\\=\frac{1}{3}\cdot \frac{4}{1}\left( {\underset{{x\to \,0}}{\mathop{{\lim }}}\frac{{\sin 4x}}{{4x}}} \right)=\frac{4}{3}\left( 1 \right)=\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}}$ | This appears to be a case of the special trigonometric limits with cos ($ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\frac{{1-\cos \,x}}{x}=0$), so try to use some algebra to figure it out. 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{1-1}{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 }}$, try turning 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)\,=\left( {\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin t}}{{2t}}\,\,} \right)\left( {\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{1}{{\cos t}}} \right)\\&=\left( {\frac{1}{2}} \right)\left( {\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin t}}{t}} \right)\left( {\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{1}{{\cos t}}} \right)=\frac{1}{2}\cdot 1\cdot \frac{1}{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$, 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}{3}} \right)\left( {\frac{2}{2}} \right)\left( {\frac{h}{h}} \right)\frac{{\sin \left( {3h} \right)}}{{\sin \left( {2h} \right)}}=\left( {\frac{3}{2}} \right)\left( {\frac{{2h}}{{3h}}} \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}\left( {\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {3h} \right)}}{{3h}}} \right)\left( {\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{2h}}{{\sin \left( {2h} \right)}}} \right)=\frac{3}{2}\cdot 1\cdot 1=\frac{3}{2}\end{align}$ |
| $ \displaystyle \underset{{t\to 0}}{\mathop{{\lim }}}t\sec t$ | 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$-value (where you have a $ y$-value) and you go to another $ x$-value 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.
Here is a real-world 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 15 pounds, for example, or any number of pounds between 7 and 20 pounds, since human weight is continuous. 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. In other words, for a weight to get from some number to another, it must have hit all the numbers in between.
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( a \right)\ne f\left( b \right)$, 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]$ | $ 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$. Thus, 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 this 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$. Thus, 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. | The function is continuous on $ [–1,5]$, so we can go on. Plug in the endpoints to see what the values of $ f\left( x \right)$ are 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 the IVT, there exists a number $ c$ in $ [–1,5]$ such that $ f\left( c \right)=0$. To find that “$ c$”, set the original function to 0, use grouping to factor, set each factor 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}$ $ x=1$ is the only solution to the function, and since $ -6<1<108$, $ c=1$. Note again that the $ \boldsymbol {c}$-value you get must be in the original interval. | |||||||||||||||
| Does the Intermediate Value Theorem guarantee a “$ 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. | The function is continuous on $ [–3,0]$, so we can go on. Plug in the endpoints to see what the values of $ f\left( x \right)$ are at those points: $ f\left( {-3} \right)={{\left( {-3} \right)}^{3}}-3\left( {-3} \right)-2=-20$ 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 –20 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?
| Note that this problem uses Compositions of Functions.
To find out if $ h\left( x \right)$ has a zero on the interval $ [2,4]$, 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$-value 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.
Limits at Infinity
Limits at Infinity exist when the $ x$-values (not the $ y$-value) go to $ \infty$ or $ -\infty $. This can happen with Rational Functions and 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 Rational Functions | Example |
| To get the end behavior asymptote (EBA), 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}}\,\,\,\,\,(=\frac{{x+2}}{{\left( {x-2} \right)\left( {x+2} \right)}})$
Notice that even though we can take out a removable discontinuity ($ x+2$, so hole at $ \left( {-2,-\frac{1}{4}} \right)$), 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$. Use long division to find this linear equation.
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 (oblique asymptote): $ 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:
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 limit problems have $ x$’s on the top and bottom, when trying to plug in $ \infty $ or $ -\infty $, we’ll typically get what we call indeterminate form – something like $ \displaystyle \frac{\infty }{\infty }$. In these cases, use the tricks of either 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}}$
| There is an EBA (end behavior asymptote) at $ y=3$, since we divide coefficients when the degree (highest exponent) is the same on the top and bottom. We can also divide each term by $ x$, since this is the highest degree (exponent of 1): $ \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}}$ | There is 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 of 1): $ \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}}}}$; we have to use $ \displaystyle -\sqrt{{{{x}^{2}}}}$, since the limit is going to $ -\infty $, so $ \displaystyle x<0$. Divide each term by either $ x$ or $ \displaystyle -\sqrt{{{{x}^{2}}}}$; use $ \displaystyle -\sqrt{{{{x}^{2}}}}$ for terms under the square roots: $ \displaystyle \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$ (Here is a graph: |
| Find the limit: $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\left( {x-\sqrt{{{{x}^{2}}+x}}} \right)$ | 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, by multiplying the numerator and denominator by the conjugate of the numerator. At the end of the second line, 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 $, and can see this by looking at a graph. So, technically speaking, no limit exists! |
| Find the limit: $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\frac{1}{{3x+\sin x}}$ | 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; a sequence is basically a function using the natural numbers as $ x$-values. Thus, 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.
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, use the same rules as Limits at Infinity above, since those limits were also going to $ \infty $. Look for the end behavior of the sequences; what our $ y$-value is converging to, when our $ x$-value 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!