Squeeze Theorem

The squeeze theorem states that if $f\left(x\right)$, $g\left(x\right)$, and $h\left(x\right)$ are functions such that $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ and $li{m}_{x\to a}f\left(x\right)=li{m}_{x\to a}h\left(x\right)=L$, then the limit $li{m}_{x\to a}g\left(x\right)=L$. In other words, one way to find the limit of a function $g\left(x\right)$ as $x$ approaches some point $a$ is by finding two functions $f\left(x\right)$ and $h\left(x\right)$ that enclose $g\left(x\right)$. If $f\left(x\right)$ and $h\left(x\right)$ come to the same limit $L$ as $x$ approaches $a$, then the $g(x)$ must equal $L$ too. Notice the two conditions of the squeeze theorem: the functions $f\left(x\right)$ and $h\left(x\right)$ need to be above and below $g\left(x\right)$, respectively, and they both need to come to the same limit value.

In choice A, the two red functions both fall below $g\left(x\right)$ in the area near $x=0$. Since the limit is taken in that region, these two red functions are not usable for the squeeze theorem.

Choice B depicts two red functions which are fully above and fully below $g\left(x\right)$, including the area near $x=0$. However, they approach two different $y$-values as $x$ approaches zero. The top function approaches $y=1$ while the bottom function approaches $y=-1$. These two functions will not allow you to find the exact value of the limit of $g\left(x\right)$ as $x$ approaches zero.

Choice C shows two red functions which are entirely above and entirely below $g\left(x\right)$, respectively. Additionally, as $x$ approaches zero, they both approach the same value, $y=0$. For these reasons, the two red functions in choice C are usable for the squeeze theorem. Further, they suggest that the limit $g\left(x\right)=0$.

Choice D shows two red functions which are above and below $g\left(x\right)$ near $x=0$. However, they are approaching different values. The top function is going to 2 while the bottom is going to –1. These two functions will not allow you to find the exact value of the limit of $g\left(x\right)$ as $x$ approaches zero.

Even though the problem doesn’t explicitly state the function $g\left(x\right)$, the squeeze theorem can help determine the limit of $g$ as $x$ approaches 3, as long as the two conditions of the theorem are met.

The squeeze theorem says that if $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ and $f(x)=h(x)=L$, then the limit $g(x)$ is also L. In this problem, the first condition, $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$, is given as true. This means to find the limit of $g\left(x\right)$ as $x$ approaches 3, check if $f\left(x\right)$ and $h\left(x\right)$ both come to the same limit at that point, and determine the value of those limits.

For this problem, the limits of $f\left(x\right)$ and $h\left(x\right)$ as $x$ approaches 3 can be determined by plugging in $x=3$.

$f\left(x\right)=(-2x^2+12x-18)$
$=-2{\left(3\right)}^2+12\left(3\right)-18$
$=-2\left(9\right)+12\left(3\right)-18$
$=-18+36–18$
$=0$

$h(x)=(\frac{1}{3}{x}^2-2x+3)$
$=\frac{1}{3}{\left(3\right)}^2-2\left(3\right)+3$
$=\frac{1}{3}\left(9\right)-2\left(3\right)+3$
$=3-6+3$
$=0$

So $f(x)=h(x)=0$. Since both limits have the same value, zero, the squeeze theorem dictates that the limit of $g\left(x\right)$ as $x$ approaches 3 must also be zero.

$g(x)=0$

It is stated that $b\left(x\right)$ lies between $a\left(x\right)$ and $c\left(x\right)$, so all that is needed to apply the squeeze theorem is determining whether $a$ and $c$ have the same limit as $x$ approaches –2. Both of those limits can be found by plugging in $x=-2$.

$a(x)=\frac{1}{4}{(-2)}^2+(-2)+3$
$=\frac{1}{4}\left(4\right)-2+3$
$=1-2+3$
$=2$
$c(x)={(-2)}^2+4(-2)+6$
$=4-8+6$
$=2$

Both $a$ and $c$ have a limit of 2 as $x$ approaches –2. Because the function $b\left(x\right)$ lies between $a\left(x\right)$ and $c\left(x\right)$, the squeeze theorem dictates that $\underset{x\to -2}{lim}b(x)$ is also 2. 

The problem states that $g\left(x\right)$ is a function lying between $f\left(x\right)$ and $h\left(x\right)$. So, the first condition of the squeeze theorem is met. The only thing necessary to check is if $f$ and $h$ come to the same limit as $x$ approaches 8. This can be checked by substituting $x=8$ into each function.

$f(x)=\sqrt{8+1}=\sqrt{9}=\pm 3$

Since we are talking about gas prices, only use the positive value for this limit.

$h(x)=\frac{1}{6}\left(8\right)+\frac{5}{3}=\frac{8}{6}+\frac{5}{3}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3$

It is now clear that $f$ and $h$ do come to the same limit as $x$ approaches 8. Therefore, the squeeze theorem can be applied to find the value of $g(x)$. That limit is the same as $f(x)$ and $h(x)$, which is 3.

To use the squeeze theorem, $r\left(x\right)$ must lie between $p\left(x\right)$ and $q\left(x\right)$, and $p$ and $q$ must have the same limit as $x$ approaches 20. In this problem, it is given that $r$ is between $p$ and $q$ when the problem states that $r$ is bounded below by $p$ and bounded above by $q$. This means the only remaining condition is checking if $p$ and $q$ have the same limit as $x$ approaches 20. This can be done by substituting $x=20$ into both functions.

$p(x)=30\left(20\right)+\mathrm{5,000}=600+\mathrm{5,000}=\mathrm{5,600}$
$q(x)=\frac{3}{4}{\left(20\right)}^2+\mathrm{5,300}=\frac{3}{4}\left(400\right)+\mathrm{5,300}=300+\mathrm{5,300}=\mathrm{5,600}$

Because $p$ and $q$ have the same limit as $x$ approaches 20, and because $r$ lies between them, the squeeze theorem dictates that $\underset{x\to 20}{lim}r(x)=\mathrm{5,600}$.