Limits and Continuity

This section covers:

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

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 x approaches 0 is 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), or it could get there, and it would still be a limit!

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 what a limit might look like graphically; note that when we factor this function (difference of cubes), we find we have a removable discontinuity, or hole:

Limits Graphically

Finding Limits Algebraically

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

Finding Limits Algebraically

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  \(\underset{{x\to {{c}^{+}}}}{\mathop{{\lim }}}\,f\left( x \right)=L\)  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:

Definition of Limits and Continuity

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 Showing Limits

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

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

Trig Function Asymptotes

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:

Continuity

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:

Limits with Sine and Cosine

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

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).

Limits of Sin Functions

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.   So the baby’s age is the x value, and the baby’s weight is the y value in this case, with the interval being between 0 and 12 months, inclusive.   This makes 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:

IVT

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

Intermediate Value Theorem Examples 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.

We learned earlier how to get the vertical asymptotes of functions, including trig functions.

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, saying these limits are \(\infty \) or \(-\infty \) is more precise). To determine which one it is, we can put in numbers (for $latex x) really close to the VA on either side (to see what direction the graph is going), or use a graphing calculator

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.

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:

Calculus One Over X

 

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 \(–2\).   

So \(\displaystyle \underset{{x\to \,-{{2}^{+}}}}{\mathop{{\lim }}}\,\frac{1}{{x+2}}=\infty \).  (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 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 \(3\). 

So we have \(\displaystyle \underset{{x\to \,{{3}^{-}}}}{\mathop{{\lim }}}\,\frac{x}{{3-x}}=\infty \).

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 \(x=3\), and a vertical asymptote at \(x=-2\):

\(\require{cancel}  \displaystyle \frac{{x-3}}{{{{x}^{2}}-x-6}}=\frac{{\cancel{{x-3}}}}{{\cancel{{\left( {x-3} \right)}}\left( {x+2} \right)}}=\frac{1}{{x+2}}\)

Now we can plug in \(3\) for \(x\) (even though the limit comes from the left), to get \(\displaystyle \frac{1}{{3+2}}=\frac{1}{5}\).

So \(\displaystyle \underset{{x\to \,{{3}^{-}}}}{\mathop{{\lim }}}\,\frac{{x-3}}{{{{x}^{2}}-x-6}}=\frac{1}{5}\).

Find the one-sided limit:

\(\displaystyle \underset{{x\to \,{{{\frac{\pi }{2}}}^{+}}}}{\mathop{{\lim }}}\,\left( {\sec \left( x \right)} \right)\)

 

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).

Calculus Sec

So we have \(\displaystyle \underset{{x\to \,{{{\frac{\pi }{2}}}^{+}}}}{\mathop{{\lim }}}\,\left( {\sec \left( x \right)} \right)=-\infty \).

 

Limits at Infinity

Limits at Infinity exist when the x values (not the y) go to  \(\infty\)  or  \(-\infty \),  so when we have rational functions.  This is because when we have rational functions,  we’re usually dealing with a horizontal asymptote (HA) (which is an end behavior asymptote, or EBA).   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:

Calculus Review of Horizontal Asymptotes

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:

Limits at Infinity

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  \(\frac{\infty }{\infty }\). 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.    

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.

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):

\(\displaystyle \underset{{x\to \,-\infty }}{\mathop{{\lim }}}\,\frac{{\frac{{2x}}{x}-\frac{1}{x}}}{{\frac{{\sqrt{{{{x}^{2}}+x-3}}}}{{-\sqrt{{{{x}^{2}}}}}}}}=\underset{{x\to \,-\infty }}{\mathop{{\lim }}}\,\frac{{2-\frac{1}{x}}}{{-\sqrt{{\frac{{{{x}^{2}}+x-3}}{{{{x}^{2}}}}}}}}=\underset{{x\to \,-\infty }}{\mathop{{\lim }}}\,\frac{{2-\frac{1}{x}}}{{-\sqrt{{1+\frac{1}{x}-\frac{3}{{{{x}^{2}}}}}}}}=\frac{{2-0}}{{-\sqrt{{1+0-0}}}}=-2\)

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!

 

Learn these rules, and practice, practice, practice!


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On to Definition of the Derivative  – you are ready!