## Is Cady Right, or Does the Limit Exist?

I recently watched the movie Mean Girls. At the end of the movie, there is a scene where the main character is shown a limit and asked to evaluate it. After quickly working out the problem in a tight competition for the academic decathlon, Cady looks up from her paper and shouts, “THE LIMIT DOES NOT EXIST!”

And of course, being the math oriented person that I am, I just was left wondering whether the limit actually exists after Mean Girls concluded. So I figured I’d share my findings.

First of all, here’s the limit shown at the end of Mean Girls: $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)}$$

## Where do we start?

Typically when I’m evaluating limits, I like to assume that the function is continuous at the point we’re evaluating the limit and just plug it in. This usually doesn’t work, but it will frequently give us a hint as to which method we can use to evaluate it. This basically just takes advantage of the basic limit properties of limits if it works out. But we need to make sure we’re not causing any problems when we do this.

So let’s just start with plugging in zero for x in the function who’s limit we’re looking for.

$$f(x) = \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)}$$ $$f(0)=\frac{ln(1-0) \ – sin(0)}{1-cos^2(0)}$$ $$f(0)=\frac{ln(1-0) \ – sin(0)}{1- \big( cos(0) \big)^2}$$ $$f(0)=\frac{0 \ – 0}{1- 1} = \frac{0}{0}$$

Turns out, this doesn’t really work this time. Because we end up with the indeterminate form of $\frac{0}{0}$. Since it’s never fine to divide by zero, this isn’t allowed.

So, we’ll need to think of another way to evaluate this limit. But like I said before, the fact that we got this indeterminate form gives us a hint of how we can evaluate this limit properly. This indeterminate form actually is one of the conditions that tells us we can evaluate this limit using L’Hospital’s Rule.

## L’Hospital’s Rule

Which means we can create a new limit. We will still have $x \to 0$ in our new limit. But we’ll take the limit of a different function. And hopefully this one will be easier to evaluate.

This new limit will actually just be made up of a fraction where the top of the fraction is just the derivative of the original numerator. And the bottom of this new faction is just the derivative of the denominator of the original fraction. So we know that $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)} \ = \ \lim_{x \to 0} \ \frac{\frac{d}{dx} \big[ ln(1-x) \ – sin(x) \big] }{\frac{d}{dx} \big[ 1-cos^2(x) \big] }$$

Remember, we are not finding the derivative of the fraction as a whole, so we should not use the quotient rule. We will need to apply the chain rule to find the derivative of a few of these terms. Doing so tells us that $$\lim_{x \to 0} \ \frac{ln(1-x) \ – sin(x)}{1-cos^2(x)} \ = \ \lim_{x \to 0} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$

## Now can we evaluate this limit?

The point of using L’Hospital’s Rule is that we should end up with a limit that’s easier to evaluate. So, let’s try the same thing we did earlier and see what happens. Since we have a limit as $x \to 0$, let’s just plug zero in for x and see what we get. $$\lim_{x \to 0} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ $$\frac{ – \ \frac{1}{1-(0)} \ – \ cos(0) }{ 2cos(0) \cdot sin(0)}$$ $$\frac{ – \ 1 \ – \ 1 }{ 2(1)(0)} = \frac{-2}{0}$$

And you can see we’ve divided by zero. So, this isn’t going to work. You can’t divide by zero, it breaks the rules of math. Therefore, we can’t just evaluate this limit by plugging in zero for x.

However, we are getting closer. The reason I say this is that we don’t have the indeterminate form of $\frac{0}{0}$ anymore. Since we have some other number divided by zero, that tells us that this limit will likely be either $\infty$ or $- \infty$.

The reason for this is that the numerator of the fraction is going toward the number -2. Whether you approach from the left or the right of x = 0, the top of our fraction is going to be a number close to -2.

Meanwhile, as x gets really, really close to 0, the denominator is going to also get really close to 0. But we don’t care about what the denominator is when x = 0. Since we’re dealing with a limit we care about the denominator when x is really close to 0. But depending on which side of zero we’re on, the denominator could be positive or negative.

If we have a fraction where the numerator is some non-zero number, and the denominator is getting infinitely close to zero, the fraction as a whole will become infinitely large. We just need to figure out if it’s positive or negative. Well, we can figure this out by splitting the limit up into two one-sided limits and see what those tell us about the two-sided limit.

## Left-Sided Limit

Let’s start with the limit where x is approaching zero from the left. $$\lim_{x \to 0^-} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ We already know that the numerator of this fraction is going to approach -2 as x approaches 0 from either side. So, let’s just look at the denominator of this fraction.

What happens to $2cos(x)sin(x)$ as $x \to 0$? Well, we already said it will approach 0, but let’s just consider what’s going on as x approaches 0 from the left. If we’re approaching 0 from the left, that means we want to consider x values that are super close to 0, but will be negative. For negative x values approaching 0, cos(x) will approach 1, and will be slightly smaller than 1. And sin(x) will approach 0, and will be negative numbers very close to 0.

This is because sin(x) is negative for x values very close to 0. We can see this looking at a graph of $f(x)=sin(x)$.

Based on this, as x approaches 0, 2cos(x)sin(x) will become the product of 2, 1, and a negative number very close to 0. The product of two positive numbers and a negative number will be a negative number. So we know that this denominator will be getting closer and closer to 0, but will be a negative number in this left sided limit.

So we can think of the fraction as a whole, as a positive number very close to 2, divided by a negative number very close to 0. As x goes to 0 from the left, this fraction will get infinitely large and will be negative because it’s a positive divided by a negative. In other words, as $x \to 0^-$, $\frac{ - \ \frac{1}{1-x} \ - \ cos(x) }{ 2cos(x) \cdot sin(x) } \to \frac{2}{-0}$. Or $$\lim_{x \to 0^-} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }=-\infty$$

## Right-Sided Limit

The right-sided limit will work out very similarly, but with one main difference. $$\lim_{x \to 0^+} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }$$ Again, the numerator of this fraction is going to approach -2 as x approaches 0 from the right. But let’s look at the denominator.

By the same process as the left-sided limit above, we can see that the denominator of this fraction is going to approach 0. But what we need to figure out is whether it is going to be approaching 0 from the negative side or the positive side. And the only term of our denominator that is going to be different approaching from the right side of 0, is sin(x).

Looking back up at the graph of $f(x)=sin(x)$ you can see that the function is positive (above the x axis) to the right of $x=0$. Therefore, the denominator of this fraction is going to be the product of three positive numbers, resulting in a positive number. And the fraction as a whole will be a positive number over a positive number. In other words, as $x \to 0^+$, $\frac{ - \ \frac{1}{1-x} \ - \ cos(x) }{ 2cos(x) \cdot sin(x) } \to \frac{2}{+0}$. Or $$\lim_{x \to 0^+} \ \frac{ – \ \frac{1}{1-x} \ – \ cos(x) }{ 2cos(x) \cdot sin(x) }=\infty$$

## What does this tell us about the two-sided limit?

Remember, for a limit to exist both one-sided limits need to exist AND they both need to be equal to each other. We just figured out that our left-sided limit is $-\infty$ and the right-sided limit is $\infty$. Since one is positive and one is negative, they clearly aren’t equal. Each one-sided limit does not give the same result. Therefore, the limit we are trying to find indeed does not exist.

So, it turns out Cady was right in the end of Mean Girls. The limit does not exist.

## Solution – Find the values of a and b that make the function differentiable everywhere.

Find all values of $a$ and $b$ that make the following function differentiable for all values of $x$.

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ ax+b & \mbox{if } x>-1 \end{cases}$$

When trying to solve a problem like this, there are actually two things you will need to consider for our function $f(x)$.  Obviously, we need to make sure that it’s differentiable everywhere, but this actually implies something else that we will want to consider as well.

Since a function being differentiable implies that it is also continuous, we also want to show that it is continuous.  The reason for this is that any function that is not continuous everywhere cannot be differentiable everywhere.  Once we make sure it’s continuous, then we can worry about whether it’s also differentiable.

## Making sure f(x) is continuous everywhere

I’m not going to go into quite as much detail to show the part about making sure the function is continuous because I have already done this, which you can see by clicking here.

To make sure $f(x)$ is continuous at $x=-1$ we need to make sure that $$\lim_{x \to -1} f(x) = f(-1).$$  Since we have a piecewise function, we will need to consider each one-sided limit, but in this case only the right sided limit will tell us something useful.

$$\lim_{x \to -1^{+}} f(x) = f(-1)$$

$$\lim_{x \to -1^{+}} ax+b = b(-1)^2-3$$

$$a(-1)+b=b-3$$

$$-a+b=b-3$$

$$-a=-3$$

$$a=3$$

So now we know that $f(x)$ will be continuous everywhere as long as $a=3$.  However, this doesn’t really tell us that $f(x)$ is differentiable everywhere as well.

## Making sure f(x) is differentiable everywhere

We now know that we will need to let $a=3$ in order for this function to be continuous and to have a chance of being differentiable.  As a result, we can say that we are now trying to make this function differentiable everywhere:

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ 3x+b & \mbox{if } x>-1 \end{cases}$$

We can see that the only place this function would possibly not be differentiable would be at $x=-1$.  The reason for this is that each function that makes up this piecewise function is a polynomial and is therefore continuous and differentiable on its entire domain.  The only place we may have a problem is when we have to switch between the two functions.

#### What does it mean for a function to be differentiable?

It means that its derivative exists for all values of $x$.  In other words, we need to be able to find its derivative no matter what $x$ is.

However, as I mentioned above, in this case we really only need to make sure that we can find the derivative of $f(x)$ when $x=-1$ since we know it would exist for all other values of $x$.  By using the definition of a derivative, we need to make sure the following limit exists at $x=-1$.

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Since we need to check this when $x=-1$, we can plug in $-1$ for $x$.  Therefore, we need to make sure this limit exists:

$$\lim_{h \to 0} \frac{f(-1+h)-f(-1)}{h}$$

I went over this limit definition in greater detail previously.  If you want a refresher on where this is coming from you can find that by clicking here.

Just like when we had to find the limit to make sure that $f(x)$ was continuous, we will need to consider each one sided limit separately in order to find this limit.  And also like when we checked for continuity, each one sided limit is going to require the use of a different section of our piecewise function.

#### Setting up the limits

When $h$ is slightly less than $0$, and we are considering the left sided limit, $f(-1+h)$ would need to be found using the $y=bx^2-3$ because this would involve inputting $x$ values which are less than $-1$.

By the same reasoning, when $h$ is slightly greater than $0$, and we are considering the right sided limit, $f(-1+h)$ would need to be found using the $y=3x+b$ because this would involve inputting $x$ values which are greater than $-1$.  Therefore, we need to consider the following one sided limits:

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

#### Now what do we do with these limits?

Now remember, as I discussed in the lesson about one-sided limits, in order for a limit to exist we need both of its one-sided limits to exist and they need to be equal.  Therefore, in order to show that the derivative of $f(x)$ exists at $x=-1$, these two one-sided limits need to be equal to each other.  Before setting them equal to each other, first we’ll simplify them a bit.  First the left side limit.

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)(-1+h)-3\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(1-2h+h^2)-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b-2bh+bh^2-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{b-2bh+bh^2-3-b+3}{h}$$

$$=\lim_{h \to 0^{-}} \frac{bh^2-2bh}{h}$$

$$=\lim_{h \to 0^{-}} \frac{h(bh-2b)}{h}$$

$$=\lim_{h \to 0^{-}} bh-2b$$

$$=-2b$$

And now the right sided limit.

$$=\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{\Big[-3+3h+b\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{-3+3h+b-b+3}{h}$$

$$=\lim_{h \to 0^{+}} \frac{3h}{h}$$

$$=\lim_{h \to 0^{+}} 3$$

$$=3$$

Now if we set these two simplified versions of the one-sided limits equal to each other, we get

$$-2b=3$$

$$b=-\frac{3}{2}$$

#### What does this tell us?

So now if we put both pieces together, we know that $a=3$ will ensure that $f(x)$ is continuous and then making $b=-\frac{3}{2}$ will also make sure $f(x)$ is differentiable at $x=-1$.  This would in turn make $f(x)$ differentiable for all values of $x$, or make it differentiable everywhere.

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## Solution – Find the values of a and b that make f continuous everywhere

Consider the following piecewise function: $$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+3 & \mbox{if } 2\leq x<3 \\ 2x-a+b & \mbox{if } x\geq 3 \end{cases}$$

Problem credit: Chapter 2.4 #36 in Single Variable Calculus: Concepts and Contexts by James Stewart.

## How to find a and b for a piecewise function to be continuous everywhere.

When we see piecewise functions like this and our goal is to make sure it is continuous everywhere, there are two cases we want to consider separately:

1. making each individual piece continuous on its reduced domain,
2. and making the different pieces line up when our function switches between them.

First let’s consider each piece individually on the reduced domain where $f$ is actually defined by that function.

Piece 1: $y=\frac{x^2-4}{x-2}$ when $x<2$

When considering a function to find x values where the function is not continuous, two very common things to look for would be taking the square root of a negative number or dividing by zero.  Obviously we don’t need to worry about taking the square root of a negative number here since we don’t have any square roots.  However, we do have a fraction, and therefore we should consider if there is an x value that would cause us to divide by zero.

Looking at the denominator of our fraction, we see that we would divide by zero when we plug in some number for x that causes

$$x-2=0.$$

Clearly, this would happen when $x=2$.  And in fact, this is the only value for x where $y=\frac{x^2-4}{x-2}$ is not continuous.  However, remember that I said we want to see if there are any x values that cause the function not to be continuous within our limited domain for the specific piece of the function.  Since $x=2$ is NOT in the domain $x<2$, we can say that this piece of our function is continuous on its entire domain and won’t contribute any discontinuities to f.

Piece 2: $y=ax^2-bx+3$ when $2\leq x<3$

Looking at this piece of our piecewise function, clearly we need to consider our constants a and b.  Since our function f is a function of x (indicated by f(x)), we can consider the other letters in this piece of our function (a and b) to be constants.  I discussed this in a bit more detail here, but it basically means that a and b are some set number, they do not change.

The important thing to realize though, is that regardless of what values we choose for a and b, this piece of our function will just be a polynomial.  More specifically, it will be a quadratic function.  No matter what our a and b end up being we will have an $x^2$ term, an x term, and a constant term.  Since we know this piece of our function will be a polynomial no matter what our a and b are we can say that this piece of our function will be continuous for any a and b we select.

This is the case because polynomials are always continuous everywhere.  If this section of our function would be continuous everywhere, it would certainly be continuous on our limited domain of $2\leq x<3$.  Therefore, we don’t need to worry about our a and b causing any discontinuities here.

Piece 3: $y=2x-a+b$ when $x\geq 3$

This is actually going to be similar to piece 2 above.  We can see by looking at this function that no matter what a and b values we select here, they will both be constants which will combine to a single constant when we combine like terms.  Once we do this, we will be left with an x term and a constant term.  Therefore, this is also a polynomial (in this case it would also be a linear function), and would be continuous everywhere as a result.  Because of this, it must be continuous on our limited domain of $x\geq 3$.

Now we have shown that we will not get any discontinuities within any of our restricted domains that make up our piecewise function, but what about on their edges.  We also have to make sure we are selecting an a and a b value so that we can switch from one part of our function to the other without jumping up or down and leaving a hole in our function.  If we do not select the correct a and b we may get discontinuities at the x values where we switch from one function to the other.

In order to prevent this, we need to make sure that our first two pieces of f(x) transition smoothly when $x=2$ (this is the place we switch between these two functions) and we need to make sure that the second and third piece of our function have a smooth transition between them at $x=3$.

Continuity at $x=2$

Remember by definition, a function f(x) is continuous at $x=a$ if $$\lim_{x \to a} f(x) = f(a).$$

We need to apply this rule here to make sure that our function f is continuous at $x=2$. Therefore, we need to pick an a and a b so that $$\lim_{x \to 2} f(x) = f(2).$$

As I explained here, a limit will only exist if both of its one-sided limits exist and are equal to each other.  Therefore, rather than finding the above limit, we actually want to find these two one-sided limits:

$$\lim_{x \to 2^{-}} f(x) = f(2)$$

$$\lim_{x \to 2^{+}} f(x) = f(2).$$

Let’s start with the left-sided limit.  Since we are approaching $x=2$ from the left, we are looking at x values that are slightly less than 2.  Therefore, we need to use the piece of our function that is defined for $x<2$.  So the limit we really want to find, and the equation we want to solve is:

$$\lim_{x \to 2^{-}} \frac{x^2-4}{x-2} = f(2).$$

When attempting to solve this limit, we see that we can not just plug in 2 for x because this would cause us to divide by zero.  As a result of this, I would first suggest simplifying the function whose limit we are trying to find.  Using the difference of squares rule on the top of our fraction we can rewrite this limit as

$$\lim_{x \to 2^{-}} \frac{(x+2)(x-2)}{x-2} = f(2).$$

Once we do this, we can see that the top and bottom of our fraction contain a $(x-2)$ term so we can use them to cancel each other out.  This leaves us with

$$\lim_{x \to 2^{-}} (x+2) = f(2).$$

$y=x+2$ is a linear function and is continuous everywhere, so we can simply plug in $x=2$ to find this limit.

$$4=f(2)$$

Now we need to figure out the right side of the equation.  To find f(2), we just need to plug in $x=2$ into our function f(x), which is our original piecewise function.  The first thing we need to do is decide which piece defines our function when $x=2$.  We know that $f(x)=ax^2-bx+3$ when $2\leq x<3$.  We can use this part to find f(2) because this is the piece that defines f when $x=2$.  Plugging in 2 for x we get

$$4=ax^2-bx+3$$

$$4=a(2)^2-b(2)+3$$

$$4=4a-2b+3$$

$$1=4a-2b$$

Notice we have only one equation at this point which relates two unknown constants, making it impossible to find one unique solution for a and b.  To do this we will need another equation relating these two constants.

I would also like to point out that this equation came from using the equation involving only the left sided limit, and we still need to look at the right sided limit.

$$\lim_{x \to 2^{+}} f(x) = f(2)$$

If we look at this limit, we see that we would be approaching $x=2$ from the right side, which would mean we are looking at x values slightly larger than 2.  For x values slightly larger than 2, but infinitely close to 2, we would use the $y=ax^2-bx+3$ piece to define f.  We already found f(2) above, so putting these two facts together we see

$$\lim_{x \to 2^{+}} ax^2-bx+3 = a(2)^2-b(2)+3$$

Since the function whose limit we are trying to find is continuous everywhere (since it’s a polynomial) we can just plug in 2 for x to find this limit.

$$a(2)^2-b(2)+3=a(2)^2-b(2)+3$$

$$4a-2b+3=4a-2b+3$$

Notice both sides of this equation are the same.  Because of this, this equation will actually be true no matter what we put in for a and b.  This doesn’t really help us at all in this case, but it was important to test it out and see what it told us.  So we know that f will be continuous at $x=2$ as long as

$$1=4a-2b.$$

At this point we have found a set of a and b values that make f continuous at $x=2$.  But we need to find just one a and one b that will accomplish this for all x values.

But this does not tell us anything about whether it would also be continuous at $x=3$.  Checking that may give us another relationship between a and b that we can use to find the single unique solution for the two constants that will make f continuous everywhere.

Continuity at $x=3$

Making sure that f is continuous at $x=3$ will be an extremely similar process to what we just did around $x=2$.  Similar to above, we will need to make sure the following equations are true:

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

$$\lim_{x \to 3^{+}} f(x) = f(3).$$

Let’s start with the right sided limit.  This means that we are getting closer and closer to $x=3$ and we are coming from x values that are slightly larger than 3.  Therefore, we will need to use the piece of our function that defines f for $x\geq 3$.

$$\lim_{x \to 3^{+}} 2x-a+b = f(3)$$

Since a and b are both constants, $y=2x-a+b$ is a linear function, and is continuous everywhere as a result.  Because of this, we can just plug 3 in for x to find this limit.

$$2(3)-a+b = f(3)$$

$$6-a+b = f(3)$$

To find f(3) we just need to plug 3 in for x into the piece of our function that defines it when $x=3$, which is the third piece of f.

$$6-a+b = 2(3)-a+b$$

$$6-a+b = 6-a+b$$

We see here another case where this equation gives us no useful information.  This is because this equality will be true no matter what we plug in for a and b.  Let’s move onto the left sided limit and see what we get there.

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

Now that our x value is approaching 3 from the left side, it is slightly smaller than 3 and slowly increasing.  Since we are considering what our function is doing around $x=3$ for x values slightly smaller than 3, we will want to use the piece of f that is defined for $2\leq x<3$.  We already found f(3) above, so we will also plug that in.

$$\lim_{x \to 3^{-}} ax^2-bx+3 = 6-a+b$$

Just like before, since we know $y=ax^2-bx+3$ is a polynomial, it is continuous everywhere and we can find this limit by plugging in $x=3$.

$$a(3)^2-b(3)+3 = 6-a+b$$

$$9a-3b+3 = 6-a+b$$

$$10a-4b=3$$

Putting it all together

Now we have another relationship that relates a and b.  Alone, it isn’t very useful, but we can take the previous equation we found from ensuring f is continuous at $x=2$ and solve the following system of equations:

$$\mbox{(1): }4a-2b=1$$

$$\mbox{(2): }10a-4b=3.$$

Now that we have two equations that relate these two unknown constants, we have enough information to solve for them.  You always need at least as many equations as you have variables (or unknown constants in this case).  There are a couple different ways to solve a system of equations like this, but I will use substitution.  First we will solve for b in equation (1), then plug that into equation (2).

$$4a-2b=1$$

$$-2b=-4a+1$$

$$\mbox{(3): }b=2a-\frac{1}{2}$$

Now we will plug this into equation (2) and solve for a.

$$10a-4\Big(2a-\frac{1}{2}\Big)=3$$

$$10a-8a+2=3$$

$$2a+2=3$$

$$2a=1$$

$$a=\frac{1}{2}$$

Now we can plug this back into our equation for b, equation (3).

$$b=2\Big(\frac{1}{2}\Big)-\frac{1}{2}$$

$$b=1-\frac{1}{2}$$

$$b=\frac{1}{2}$$

Now we have shown that f(x) will be continuous at $x=2$ and at $x=3$ if $a=\frac{1}{2}$ and $b=\frac{1}{2}$.  We already showed that f is continuous for $x<2$, $2, and $x>3$ for all values of a and b.  Now we also know that f is continuous at $x=2$ and $x=3$ if $a=\frac{1}{2}$ and $b=\frac{1}{2}$.  Therefore, we can say that f would be continuous everywhere if $a=\frac{1}{2}$ and $b=\frac{1}{2}$.

In fact, we can see that f is continuous by plugging in the values we found for a and b and graphing it using Desmos.

This type of problem is one that you will likely run into a few times.  It actually comes in handy when you are trying to solve a similar type of problem that requires you to make a function differentiable everywhere.  If you would like to see this I have worked through one of these and you can see that by clicking this link: Solution – Find the values of a and b that make the function differentiable everywhere.

## SHORTCUT – More examples

Check out the other topics I’ve covered and the problems I’ve worked through.  You can see a list of related lessons on my limits page.

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## One-sided Limits

To end the last lesson, Limits – Intro, I mentioned that a limit would not exist if you did not approach the same $y$ value as you approach a given $x$ value from both the left and right.  For example, consider $f(x)$ shown in Figure 1.2 again (shown again below), but this time let’s find:

$$\lim_{x \to -1}f(x).$$

In the previous section I mentioned approaching the given $x$ value from the left and the right side to find the limit. This is necessary when finding two-sided limits (usually just referred to as “limits”), but let’s consider each side separately. In other words, we will consider what $y$ we get close to as we approach $x=-1$ from the left side as one problem and from the right side as a separate problem.

## Approaching from the left side

First we will approach $x=-1$ from the left side. This is denoted like this:

$$\lim_{x \to -1 ^{-}}f(x).$$

Notice the little $^-$ to the right of the $-1$. This tells us the we are approaching $x=-1$ only from the negative side, or the left side. As shown below in Figure 1.3, as we approach $x=-1$ from the left side, we get closer to $y=2$.

Therefore, we can say that

$$\lim_{x \to -1 ^{-}}f(x) = 2.$$

## Approaching from the right side

Similarly, if we only consider what $y$ value we approach as we get close to $x=-1$ from the right side, or the positive side, we are finding:

$$\lim_{x \to -1 ^{+}}f(x).$$

Just like before, we only want to consider approaching this specific $x$ value from one side:

As you can see, if we start on this function to the right of $x=-1$ and we move toward $x=-1$ along the function, we get closer and closer to a $y$ value of $4$. Therefore, we can say that this one-sided limit has a value of $4$, or

$$\lim_{x \to -1 ^{+}}f(x) = 4.$$

## Putting them together

Now, we have found both one sided limits of this function around $x=-1$. As we approach $x=-1$ from the left side, we get closer to $y=2$, but as we approach $x=-1$ from the right side, we get closer to $y=4$. Since we get infinitely close to two different $y$ values depending on whether we approach $x=-1$ from the left side versus the right side, this two-sided limit actually does not exist. So,

$$\lim_{x \to -1}f(x) \ \ DNE.$$

#### Why do we need to consider each one sided limit separately?

This is an important thing to remember. In order to find any two-sided limit, you will instead find each one sided limit. If both one-sided limits are the same, then the two-sided limit will also be that same value. However, if the one-sided limits are different, the two-sided limit does not exist. In other words, $$if \ \lim_{x \to a ^{-}}f(x) = \lim_{x \to a ^{+}}f(x) = b, \ then \ \lim_{x \to a}f(x) = b.$$ $$And \ if \ \lim_{x \to a ^{-}}f(x) \neq \lim_{x \to a ^{+}}f(x), \ then \ \lim_{x \to a}f(x) \ does \ not \ exist.$$

Another method that can be used to evaluate one-sided limits if you don’t have a graph of the function available is using the properties of limits. You can learn more about the limit properties here.

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For more on limits go check out my limits page.  There’s a list of lessons and practice problems there all about limits.  Take a look to get some more practice with limits.  If you have a specific topic or problem you’re looking for and can’t find it, then email me at jakesmathlessons@gmail.com.  Send me your questions and I’ll be sure to point you in the right direction to get them answered.  I may even write a lesson and post it to make sure your question gets answered!