Computing Limits

In the previous section we saw that there is a large class of functions that allows us to use direct substitution \[\lim\limits_{x\to a}f(x) = f(a)\] to compute limits. However, there are also many limits for which direct substitution won’t work immediately. The purpose of this section is to develop techniques for dealing with some of these limits when direct substitution does not yield a direct value.

Let’s first go back and take a look at one of the first limits that we looked at and compute its exact value and verify our guess for the limit.

Before moving on from this example, let’s discuss the fact that directly substituting $x = 2$ did not work when we first tried, but it did work after some helpful algebraic simplification.

The main property we used here is that $\dfrac{(x-2)(x+6)}{x(x-2)}$ and $\dfrac{x+6}x$ are equal everywhere except at $x = 2$. In other words, they are almost the same function, but the domain of $\dfrac{x+6}x$ includes $x = 2$ while the domain of $\dfrac{(x-2)(x+6)}{x(x-2)}$ does not. Because the limit is only concerned with what happens near $x =2 $, and not at $x = 2$, both limits will be equal. Here is what each of those functions look like.

$$f(x) = \dfrac{(x-2)(x+6)}{x(x-2)}$$

$$g(x) = \dfrac{(x+6)}{x}$$

Remember, limits don't care what happens when $x = a$, only what happens for $x$-values near $a$.

In essence, we factored out the part of the numerator and denominator that was equal to 0 when $x = 2$, that is, we factored out $(x-2)$. By factoring out the root that was causing the $0$ in the numerator and the denominator, we could then find a algebraically sound way to eliminate the parts causing the limit to be become $0/0$ when we initially tried plugging in $x = 2$.

Let's add this limit property to our list from the previous section.

We will use this property frequently when simplifying limits.

There are many more kinds of indeterminate forms, where we can't even immediately tell if the value of the limit is a real number or not, and we will be discussing indeterminate forms at length in the next chapter.

Let’s take a look at a couple of more examples.

Try out these three algebraic techniques for simplifying $0/0$ limits with the following practice problems yourself.

To recap, we’ve taken a look at a few limits in which direct substitution gave the indeterminate form $0/0$ and we now have a few algebraic techniques to try in these cases.

Let’s take a look at another kind of problem that can arise in computing some limits involving piecewise functions. Note that piecewise functions are not included in the list of functions where direct substitution is guaranteed to work from the end of the limit properties section, because the function can sometimes jump from one value to another when its formula changes.

So sometimes limits of piecewise functions don't exist at the endpoint of the domain of one of its pieces, but it is quite simple to construct examples where the limits do exist. Here's a slightly modified version of the second part of the above problem to illustrate this.

Squeeze Theorem

There is one more limit technique we can use that we'll cover this section. Before we introduce it, we will need a new fact about limits that will help us to do this.

The intuition here is quite straightforward: if the output values of $f(x)$ never exceed those of $g(x)$, then the same is true of the corresponding limits since those output values are precisely what determine the value of the limit.

Note that we excluded $x = c$ when requiring $f(x) \leq g(x)$, because the limits also ignore what happens at $x = c$. This will be particularly useful for us in the case where either $f(c)$ or $g(c)$ doesn't exist, but the limit still exists.

This fact is the basis for the following theorem.

This time we are comparing three functions, but the idea is largely the same as before. Since $g(x) \geq f(x)$ and $\lim\limits_{x\to c}f(x) = L$, we know that $\lim\limits_{x\to c}g(x)$ must be at least as big as $L$. Likewise, since $g(x) \geq h(x)$ and $\lim\limits_{x\to c}h(x) = L$, we know that $\lim\limits_{x\to c}g(x)$ can be no bigger than $L$. But the only real number both as big as $L$ and no bigger than $L$ is $L$ itself, so $\lim\limits_{x\to c}g(x)$ must also equal $L$.

Here's a graph example of the Squeeze Theorem.

This is a graph of three functions. The highest function $h(x)$ and the lowest function is $f(x)$, while $g(x)$ is in between these two. All three functions approach the same output value $y = L$ at $x=c$.

Since the values of $g(x)$ must be between $f(x)$ and $h(x)$, the graph of $g(x)$ is being "squeezed" between the graphs of $f(x)$ and $h(x)$. When $f(x)$ and $h(x)$ both approach $y = L$ as $x \to c$, they likewise squeeze $g(x)$ to the same value of $y = L$.

Sometimes the Squeeze Theorem is also called the Sandwich Theorem.

Let's take a look at a classic example where the Squeeze Theorem is needed to evaluate a limit.

Evaluate the limit $\lim\limits_{x\to0}x^2\cos\left(\dfrac1x\right)$.

Click for solution

If we try direct substitution, we get a value of $0^2\cos(1/0)$, but the division by zero inside of cosine causes this method to fail. Likewise none of the algebra techniques we've reviewed this section can do anything to help us simplify the limit.

Before we can use the Squeeze Theorem, we have to identify what three functions will take the role of $f(x)$, $g(x)$, and $h(x)$. We want $x^2\cos(1/x)$ to be our $g(x)$ since the Squeeze Theorem will allow us to conclude a value for $\lim\limits_{x\to0} x^2\cos\left(\dfrac1x\right)$, assuming the limits for $f(x)$ and $g(x)$ are known and equal.

Therefore we are going to start with the fact that \[-1 \leq \cos(\theta)\leq 1\] for any angle $\theta$ we put into cosine. In particular, let's plug in $\theta = 1/x$. Then \[-1\leq\cos\left(\dfrac1x\right)\leq1.\] Notice that this inequality is not true when $x = 0$ since we would have a division by $0$ error, but that's ok since the limit (and hence the Squeeze Theorem) ignores what happens at $x = 0$.

We almost have $x^2\cos(1/x)$ in the middle, the only part we are missing is the $x^2$. So let's multiply the inequality through by $x^2$ to get \[-x^2 \leq x^2 \cos\left(\dfrac1x\right) \leq x^2.\]

Thus the middle function is now the function we care about, $x^2\cos(1/x)$, and we are comparing it to two much simpler functions above and below it, $x^2$ and $-x^2$, which will be our $h(x)$ and $f(x)$ respectively.

Lastly, before we can use the Squeeze Theorem, we have to verify the limits as $x\to0$ of the two bounding functions. \[\lim\limits_{x\to0}x^2 = 0 \hspace{4em} \lim\limits_{x\to0}-x^2 = 0\] Since these two limits are both equal to $0$, we can now invoke the Squeeze Theorem and conclude \[\lim\limits_{x\to0}x^2\cos\left(\dfrac1x\right) = 0.\] Here is a graph of the situation.

Practice Problems

Evaluate each limit, if it exists.

1. $\lim\limits_{x\to2}(8-3x+12x^2)$
Click for solution


We know that the first thing that we should try to do is simply plug in the value and see if we can compute the limit.

$$\mathop {\lim }\limits_{x \to 2} \left( {8 - 3x + 12{x^2}} \right) = 8 - 3\left( 2 \right) + 12\left( 4 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{50}}$$

Success!



2. $\displaystyle \mathop {\lim }\limits_{t \to \, - 3} \frac{{6 + 4t}}{{{t^2} + 1}}$
Click for solution


We know that the first thing that we should try to do is simply plug in the value and see if we can compute the limit.

$$\mathop {\lim }\limits_{t \to - 3} \frac{{6 + 4t}}{{{t^2} + 1}} = \frac{{ - 6}}{{10}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{3}{5}}}$$

Success!



3. $\displaystyle \mathop {\lim }\limits_{x \to \, - 5} \frac{{{x^2} - 25}}{{{x^2} + 2x - 15}}$
Click for solution


Recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Remember that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. All we need to do here is some simplification and then we’ll reach a point where we can plug in the value.

$$\mathop {\lim }\limits_{x \to \, - 5} \frac{{{x^2} - 25}}{{{x^2} + 2x - 15}} = \mathop {\lim }\limits_{x \to \, - 5} \frac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{\left( {x - 3} \right)\left( {x + 5} \right)}} = \mathop {\lim }\limits_{x \to \, - 5} \frac{{x - 5}}{{x - 3}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{4}}}$$


4. $\displaystyle \mathop {\lim }\limits_{z \to 8} \frac{{2{z^2} - 17z + 8}}{{8 - z}}$
Click for solution


Recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Remember that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. All we need to do here is some simplification and then we’ll reach a point where we can plug in the value.

$$\mathop {\lim }\limits_{z \to 8} \frac{{2{z^2} - 17z + 8}}{{8 - z}} = \mathop {\lim }\limits_{z \to 8} \frac{{\left( {2z - 1} \right)\left( {z - 8} \right)}}{{ - \left( {z - 8} \right)}} = \mathop {\lim }\limits_{z \to 8} \frac{{2z - 1}}{{ - 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 15}}$$


5. $\displaystyle \mathop {\lim }\limits_{y \to 7} \frac{{{y^2} - 4y - 21}}{{3{y^2} - 17y - 28}}$
Click for solution


Recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Remember that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. All we need to do here is some simplification and then we’ll reach a point where we can plug in the value.

$$\mathop {\lim }\limits_{y \to 7} \frac{{{y^2} - 4y - 21}}{{3{y^2} - 17y - 28}} = \mathop {\lim }\limits_{y \to 7} \frac{{\left( {y - 7} \right)\left( {y + 3} \right)}}{{\left( {3y + 4} \right)\left( {y - 7} \right)}} = \mathop {\lim }\limits_{y \to 7} \frac{{y + 3}}{{3y + 4}} = \frac{{10}}{{25}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{2}{5}}}$$


6. $\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {6 + h} \right)}^2} - 36}}{h}$
Click for solution


Recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Remember that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. All we need to do here is some simplification and then we’ll reach a point where we can plug in the value.

$$\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {6 + h} \right)}^2} - 36}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{36 + 12h + {h^2} - 36}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {12 + h} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {12 + h} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{12}}$$


7. $\displaystyle \mathop {\lim }\limits_{z \to 4} \frac{{\sqrt z - 2}}{{z - 4}}$
Click for solution


Recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Remember that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. All we need to do here is some simplification and then we’ll reach a point where we can plug in the value.

$$\mathop {\lim }\limits_{z \to 4} \frac{{\sqrt z - 2}}{{z - 4}} = \mathop {\lim }\limits_{z \to 4} \frac{{\left( {\sqrt z - 2} \right)}}{{\left( {z - 4} \right)}}\frac{{\left( {\sqrt z + 2} \right)}}{{\left( {\sqrt z + 2} \right)}} = \mathop {\lim }\limits_{z \to 4} \frac{{z - 4}}{{\left( {z - 4} \right)\left( {\sqrt z + 2} \right)}} = \mathop {\lim }\limits_{z \to 4} \frac{1}{{\sqrt z + 2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{4}}}$$


8. $\displaystyle \mathop {\lim }\limits_{x \to \, - 3} \frac{{\sqrt {2x + 22} - 4}}{{x + 3}}$
Click for solution


Recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Remember that this DOES NOT mean that the limit doesn’t exist. Simply factoring will not do us much good here so in this case it looks like we’ll need to rationalize the numerator.

$$\begin{align*}\mathop {\lim }\limits_{x \to \, - 3} \frac{{\sqrt {2x + 22} - 4}}{{x + 3}} & = \mathop {\lim }\limits_{x \to \, - 3} \frac{{\left( {\sqrt {2x + 22} - 4} \right)}}{{\left( {x + 3} \right)}}\cdot\frac{{\left( {\sqrt {2x + 22} + 4} \right)}}{{\left( {\sqrt {2x + 22} + 4} \right)}} = \mathop {\lim }\limits_{x \to \, - 3} \frac{{2x + 22 - 16}}{{\left( {x + 3} \right)\left( {\sqrt {2x + 22} + 4} \right)}}\\ & = \mathop {\lim }\limits_{x \to \, - 3} \frac{{2\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {\sqrt {2x + 22} + 4} \right)}} = \mathop {\lim }\limits_{x \to \, - 3} \frac{2}{{\sqrt {2x + 22} + 4}} = \frac{2}{8} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{4}}}\end{align*}$$


9. $\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{x}{{3 - \sqrt {x + 9} }}$
Click for solution


Recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Remember that this DOES NOT mean that the limit doesn’t exist. Simply factoring will not do us much good here so in this case it looks like we’ll need to rationalize the numerator.

$$\begin{align*}\mathop {\lim }\limits_{x \to 0} \frac{x}{{3 - \sqrt {x + 9} }} & = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\left( {3 - \sqrt {x + 9} } \right)}}\cdot\frac{{\left( {3 + \sqrt {x + 9} } \right)}}{{\left( {3 + \sqrt {x + 9} } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {3 + \sqrt {x + 9} } \right)}}{{9 - \left( {x + 9} \right)}}\\ & = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {3 + \sqrt {x + 9} } \right)}}{{ - x}} = \mathop {\lim }\limits_{x \to 0} \frac{{3 + \sqrt {x + 9} }}{{ - 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 6}}\end{align*}$$


10. Given the function $$f\left( x \right) =\begin{cases} 7 - 4x & x\lt 1 \\ x^2 + 2 & x \geq 1\\ \end{cases}$$ evaluate the following limits, if they exist.
1. $\lim\limits_{x\to-6}f(x)$
2. $\lim\limits_{x\to1}f(x)$
Click for solution


a. For this part we know that $-6\lt 1$ and so there will be values of x on both sides of -6 in the range $x \lt 1$ and so we can assume that, in the limit, we will have $x\lt 1$. This will allow us to use the piece of the function in that range and then just use standard limit techniques to compute the limit. $$\mathop {\lim }\limits_{x \to \, - 6} f\left( x \right) = \mathop {\lim }\limits_{x \to \, - 6} \left( {7 - 4x} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{31}}$$




---




b. This part is going to be different from the previous part. We are looking at the limit at $x=1$ and that is the “cut–off” point in the piecewise functions. Recall from the discussion in the section, that this means that we are going to have to look at the two one sided limits.

$$\mathop {\lim }\limits_{x \to {1^{\, - }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, - }}} \left( {7 - 4x} \right) = \underline 3 \hspace{0.25in}\hspace{0.25in}{\mbox{because }}x \to {1^ - }{\mbox{ implies that }}x \lt 1$$ $$\mathop {\lim }\limits_{x \to {1^{\, + }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, + }}} \left( {{x^2} + 2} \right) = \underline 3 \hspace{0.25in}\hspace{0.25in}{\mbox{because }}x \to {1^ + }{\mbox{ implies that }}x \gt 1$$

So, in this case, we can see that, $$\mathop {\lim }\limits_{x \to {1^{\, - }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, + }}} f\left( x \right) = 3$$ and so we know that the overall limit must exist and, $$\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \require{bbox} \bbox[2pt,border:1px solid black]{3}.$$



11. Given $$h(z) = \begin{cases} 6z & z \leq -4 \\ 1 - 9z & z \gt -4\\ \end{cases}$$ evaluate the following limits, if they exist.
    1. $\lim\limits_{z\to 7}h(z)$
    2. $\lim\limits_{z\to-4}h(z)$
    Click for solution
    

    a. For this part we know that $7\gt -4$ and so there will be values of $z$ on both sides of 7 in the range $z \gt -4$ and so we can assume that, in the limit, we will have $z \gt -4$. This will allow us to use the piece of the function in that range and then just use standard limit techniques to compute the limit.

    $$\mathop {\lim }\limits_{z \to 7} h\left( z \right) = \mathop {\lim }\limits_{z \to 7} \left( {1 - 9z} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 62}}$$
    

    ---

    
    

    b. This part is going to be different from the previous part. We are looking at the limit at $z=-4$ and that is the “cut–off” point in the piecewise functions. Recall from the discussion in the section, that this means that we are going to have to look at the two one sided limits.

    $$\mathop {\lim }\limits_{z \to \, - {4^{\, - }}} h\left( z \right) = \mathop {\lim }\limits_{z \to \, - {4^{\, - }}} 6z = \underline { - 24} \hspace{0.25in}\hspace{0.25in}{\mbox{because }}z \to - {4^ - }{\mbox{ implies that }}z \lt - 4$$ $$\mathop {\lim }\limits_{z \to \, - {4^{\, + }}} h\left( z \right) = \mathop {\lim }\limits_{z \to \, - {4^{\, + }}} \left( {1 - 9z} \right) = \underline {37} \hspace{0.25in}{\mbox{because }}z \to - {4^ + }{\mbox{ implies that }}z \gt - 4$$

    So, in this case, we can see that, $$\mathop {\lim }\limits_{z \to \, - {4^{\, - }}} h\left( z \right) = - 24 \ne 37 = \mathop {\lim }\limits_{z \to \, - {4^{\, + }}} h\left( z \right)$$ and so we know that the overall limit does not exist.

    
    
12. Evaluate $\lim\limits_{x\to5}(10+|x-5|)$, if it exists.
Click for solution


Recall the definition of the absolute value function. $$|p| = \begin{cases} p & p\geq 0\\ -p & p\lt 0 \\ \end{cases}.$$ So, because the function inside the absolute value is zero at $x=5$, we can see that, $$|x-5| = \begin{cases} x - 5 & x\geq 5\\ -(x-5) & x \lt 5\\ \end{cases}.$$

This means that we are being asked to compute the limit at the “cut–off” point in a piecewise function and so, as we saw in this section, we’ll need to look at two one-sided limits in order to determine if this limit exists (and its value if it does exist).

$$\mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {10 + \left| {x - 5} \right|} \right) = \mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {10 - \left( {x - 5} \right)} \right) = \mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {15 - x} \right) = 10\hspace{0.25in}{\mbox{recall }}x \to {5^ - }{\mbox{ implies }}x \lt 5$$ $$\mathop {\lim }\limits_{x \to {5^{\, + }}} \left( {10 + \left| {x - 5} \right|} \right) = \mathop {\lim }\limits_{x \to {5^{\, + }}} \left( {10 + \left( {x - 5} \right)} \right) = \mathop {\lim }\limits_{x \to {5^{\, + }}} \left( {5 + x} \right) = 10\hspace{0.25in}{\mbox{recall }}x \to {5^ + }{\mbox{ implies }}x \gt 5$$

So, for this problem, we can see that, $$\mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {10 + \left| {x - 5} \right|} \right) = \mathop {\lim }\limits_{x \to 5 + } \left( {10 + \left| {x - 5} \right|} \right) = 10$$ and so the overall limit must exist and, $$\mathop {\lim }\limits_{x \to {5}} \left( {10 + \left| {x - 5} \right|} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{10}}.$$



13. Evaluate $\lim\limits_{t\to-1}\frac{t+1}{|t+1|}$, if it exists.
Click for solution


Recall the definition of the absolute value function. $$|p| = \begin{cases} p & p\geq 0\\ -p & p\lt 0 \\ \end{cases}.$$ So, because the function inside the absolute value is zero at $t=-1$, we can see that, $$|t+1| = \begin{cases} t+1& t\geq -1\\ -(t+1) & t \lt -1\\ \end{cases}.$$

This means that we are being asked to compute the limit at the “cut–off” point in a piecewise function and so, as we saw in this section, we’ll need to look at two one-sided limits in order to determine if this limit exists (and its value if it does exist).

$$\mathop {\lim }\limits_{t \to \, - {1^{\, - }}} \frac{{t + 1}}{{\left| {t + 1} \right|}} = \mathop {\lim }\limits_{t \to \, - {1^{\, - }}} \frac{{t + 1}}{{ - \left( {t + 1} \right)}} = \mathop {\lim }\limits_{t \to \, - {1^{\, - }}} - 1 = - 1\hspace{0.25in}{\mbox{recall }}t \to - {1^ - }{\mbox{ implies }}t \lt - 1$$ $$\mathop {\lim }\limits_{t \to \, - {1^{\, + }}} \frac{{t + 1}}{{\left| {t + 1} \right|}} = \mathop {\lim }\limits_{t \to \, - {1^{\, + }}} \frac{{t + 1}}{{t + 1}} = \mathop {\lim }\limits_{t \to \, - {1^{\, + }}} 1 = 1\hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mbox{recall }}t \to - {1^ + }{\mbox{ implies }}t \gt - 1$$

So, for this problem, we can see that, $$\mathop {\lim }\limits_{t \to \, - {1^{\, - }}} \frac{{t + 1}}{{\left| {t + 1} \right|}} = - 1 \ne 1 = \mathop {\lim }\limits_{t \to \, - {1^{\, + }}} \frac{{t + 1}}{{\left| {t + 1} \right|}}$$ and so the overall limit does not exist.



14. Given that $7x \leq f(x) \leq 3x^2+2$ for all $x$, determine the value of $\lim\limits_{x\to 2}f(x)$.
Click for solution


This problem is set up to use the Squeeze Theorem. First, we already know that $f(x)$ is always between two other functions. Now all that we need to do is verify that the two “outer” functions have the same limit at $x=2$ and if they do we can use the Squeeze Theorem to get the answer.

$$\mathop {\lim }\limits_{x \to 2} 7x = 14\hspace{0.75in}\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 2} \right) = 14$$

So, we have, $$\mathop {\lim }\limits_{x \to 2} 7x = \mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 2} \right) = 14$$ and so by the Squeeze Theorem we must also have, $$\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 14.$$



15. Use the Squeeze Theorem to determine the value of $\lim\limits_{x\to 0} x^4 \sin\left(\frac{\pi}{x}\right)$.
Click for solution


We first need to determine lower/upper functions. We’ll start off by acknowledging that provided $x\neq 0$ (which we know it won’t be because we are looking at the limit as $x\to0$) we will have, $$- 1 \le \sin \left( {\frac{\pi }{x}} \right) \le 1.$$

Now, simply multiply through this by $x^4$ to get, $$- {x^4} \le {x^4}\sin \left( {\frac{\pi }{x}} \right) \le {x^4}.$$

Before proceeding note that we can only do this because we know that $x^4\gt 0$ for $x\neq 0$. Recall that if we multiply through an inequality by a negative number we would have had to switch the inequality signs. So, for instance, had we multiplied through by $x^3$, we would have had issues because this is positive if $x\gt0$ and negative if $x\lt 0$.

Now, let’s get back to the problem. We have a set of lower/upper functions and clearly, $$\mathop {\lim }\limits_{x \to 0} {x^4} = \mathop {\lim }\limits_{x \to 0} \left( { - {x^4}} \right) = 0$$ Therefore, by the Squeeze Theorem we must have, $$\mathop {\lim }\limits_{x \to 0} {x^4}\sin \left( {\frac{\pi }{x}} \right) = 0.$$

Assignment Problems

$\lim\limits_{x\to-9}(1-4x^3)$
$\lim\limits_{y\to 1}(6y^4-7y^3+12y+25)$
$\lim\limits_{t\to 0}\frac{t^2+6}{t^2-3}$
$\lim\limits_{z\to4}\frac{6z}{2+3z^2}$
$\lim\limits_{w\to-2}\frac{w+2}{w^2-6w-16}$
$\lim\limits_{t\to-5}\frac{t^2+6t+5}{t^2+2t-15}$
$\lim\limits_{x\to3}\frac{5x^2-16x+3}{9-x^2}$
$\lim\limits_{z\to 1}\frac{10-9z-z^2}{3z^2+4z-7}$
$\lim\limits_{x\to-2}\frac{x^3+8}{x^2+8x+12}$
$\lim\limits_{t\to8}\frac{t(t-5)-24}{t^2-8t}$
$\lim\limits_{w\to-4}\frac{w^2-16}{(w-2)(w+3)-6}$
$\lim\limits_{h\to 0}\frac{(2+h)^3-8}{h}$
$\lim\limits_{h\to 0}\frac{(1+h)^4-1}{h}$
$\lim\limits_{t\to 25}\frac{5-\sqrt{t}}{t-25}$
$\lim\limits_{x\to 2}\frac{x-2}{\sqrt{2}-\sqrt{x}}$
$\lim\limits_{z\to 6}\frac{z-6}{\sqrt{3z-2}-4}$
$\lim\limits_{z\to-2}\frac{3-\sqrt{1-4z}}{2z+4}$
$\lim\limits_{t\to 3}\frac{3-t}{\sqrt{t+1}-\sqrt{5t-11}}$
$\lim\limits_{x\to 7}\frac{\frac17 - \frac{1}{x}}{x-7}$
$\lim\limits_{y\to-1}\frac{\frac{1}{4+3y}+\frac{1}{y}}{y+1}$


Given the function $$f(x) = \begin{cases} 15 & x \lt -4 \\ 6-2x & x\geq -4\\ \end{cases}$$ evaluate the following limits, if they exist.
1. $\lim\limits_{x\to -7}f(x)$
2. $\lim\limits_{x\to-4}f(x)$
Given the function $$g(t) = \begin{cases} t^2-t^3 & t \lt 2 \\ 5t-14 & t\geq 2\\ \end{cases}$$ evaluate the following limits, if they exist.
1. $\lim\limits_{t\to -3}g(t)$
2. $\lim\limits_{t\to2}g(t)$
Given the function $$h(w) = \begin{cases} 2w^2 & w \leq 6 \\ w-8 & w \gt 6\\ \end{cases}$$ evaluate the following limits, if they exist.
1. $\lim\limits_{w\to6}h(w)$
2. $\lim\limits_{w\to2}h(w)$
Given the function $$g(x) = \begin{cases} 5x+24 & x \lt -3 \\ x^2 & -3 \leq x \lt 4\\ 1-2x & x\geq 4\\ \end{cases}$$ evaluate the following limits, if they exist.
1. $\lim\limits_{x\to -3}g(x)$
2. $\lim\limits_{x\to0}g(x)$
3. $\lim\limits_{x\to4}g(x)$
4. $\lim\limits_{x\to12}g(x)$

For the following problems, evaluate the limit, if it exists.

25. $\lim\limits_{t\to-10}(|t+10|+3)$
26. $\lim\limits_{x\to4}(9+|8-2x|)$
27. $\lim\limits_{h\to 0}\frac{|h|}{h}$
28. $\lim\limits_{t\to 2}\frac{2-t}{|t-2|}$
29. $\lim\limits_{w\to-5}\frac{|2w+10|}{w+5}$
30. $\lim\limits_{x\to 4}\frac{|x-4|}{x^2-16}$


31. Given that $3+2x \leq f(x) \leq x - 1$ for all $x$, determine the value of $\lim\limits_{x\to-4}f(x)$.
32. Given that $\sqrt{x+7} \leq f(x) \leq \frac{x-1}{2}$ for all $x$, determine the value of $\lim\limits_{x\to9}f(x)$.
33. Use the Squeeze Theorem to determine the value of $\lim\limits_{x\to 0}x^4\cos\left(\frac{3}{x}\right)$.
34. Use the Squeeze Theorem to determine the value of $\lim\limits_{x\to 0}x\cos\left(\frac{3}{x}\right)$.
35. Use the Squeeze Theorem to determine the value of $\lim\limits_{x\to 1}(x-1)^2\cos\left(\frac{1}{x-1}\right)$.