# Calculus: Limits

## Squeeze Theorem

The Squeeze Theorem is a useful tool for solving limits indirectly. The key maneuver is to figure out how to meet the requirements of the theorem. Since the theorem applies to possible situations that meet the criteria, it therefore must apply to the particular one you might be trying to solve. Presto - you have you answer.

The Squeeze Theorem: Let $I$ be an interval containing a point $a$. Also let $f$, $g$, and $h$ be functions defined on $I$, except possibly at $a$. If for every $x \neq a \in I$ it is true that

$$f(x) \leq g(x) \leq h(x)$$

and that

$$\lim\limits_{x \rightarrow a} f(x) = \lim\limits_{x \rightarrow a}h(x) = P$$

for some value $P$. Then

$$\lim\limits_{x \rightarrow a} g(x) = P$$

Stated less rigorously, if the function you're studying can be bounded above and below by two functions whose limits you do know, and those limits turn out to be the same, then because the original function is "squeezed" in between those bounding functions, its limit must be equal to that of theirs.

## Problems

1. Consider a function $f(x)$. Let $h(x)$ be an upper bound for $f(x)$ and $g(x)$ be a lower bound for $f(x)$.

Given that $\lim\limits_{x \rightarrow a} \, h(x) = b$ and $\lim\limits_{x \rightarrow a} \, g(x) = b$, what can you conclude about $\lim\limits_{x \rightarrow a} \, f(x)$?

$\lim\limits_{x \rightarrow a} f(x) = b$ follows directly from the Squeeze Theorem.

2. Let $f(x) = x^2\cos(x)$.

1. Find an upper bound and a lower bound for $f(x)$.
2. Use the bounds to show that by the Squeeze Theorem, $\lim\limits_{x \rightarrow 0} \, f(x) = 0$.
1. Notice that the range of $\cos(x)$ is $[-1,1]$. An upper bound for $\cos(x)$ is thus $1$, and a lower bound is $-1$. We can substitute these bounds into $f(x)$ to produce an upper bound $f^{+}(x)$ and a lower bound $f^{-1}(x)$:

• $f^{+}(x) = x^2 \cdot 1 = x^2$
• $f^{-}(x) = x^2 \cdot (-1) = -x^2$
2. Clearly $\lim\limits_{x \rightarrow 0} \, f^+(x) = 0$ and $\lim\limits_{x \rightarrow 0} \, f^-(x) = 0$. Thus by the Squeeze Theorem, $\lim\limits_{x \rightarrow 0} \, f(x) = 0$.

3. Let $f(x) = e^{-x}\cos(x) + 2^{-x}\sin(x)$.

1. Find an upper bound and a lower bound for $f(x)$.
2. Use the bounds to find $\lim\limits_{x \rightarrow \infty} \, f(x)$.
1. Notice that the range of both $\sin(x)$ and $\cos(x)$ is $[-1,1]$. An upper bound for both is thus $1$, and a lower bound is $-1$. We can substitute these bounds into $f(x)$ to produce an upper bound $f^{+}(x)$ and a lower bound $f^{-1}(x)$:

• $f^+(x) = e^{-x}\cdot 1 + 2^{-x}\cdot1 = e^{-x} + 2^{-x}$
• $f^-(x) = e^{-x}\cdot(-1) + 2^{-x}\cdot(-1) = -e^{-x} - 2^{-x}$
2. Clearly $\lim\limits_{x \rightarrow \infty} \, f^+(x) = 0$ and $\lim\limits_{x \rightarrow \infty} \, f^-(x) = 0$. Thus by the Squeeze Theorem, $\lim\limits_{x \rightarrow \infty} \, f(x) = 0$.

4. Let the Dirichlet function be denoted by $\delta(x) = \left\{ \begin{array}[ll] \phantom{}1 & x \text{ is rational} \\ 0 & x \text{ is irrational} \\ \end{array} \right. \\$

Let $f(x) = x^3\delta(x) - 4x\delta(x)$.

Use the Squeeze Theorem to find $\lim\limits_{x \rightarrow -2} \, f(x)$.

Hint: You must find sets of bounds for both $\lim\limits_{x \rightarrow -2^+} \, f(x)$ and $\lim\limits_{x \rightarrow -2^-} \, f(x)$.

Clearly $1$ is an upper bound for $\delta(x)$ and $0$ is a lower bound. Thus

• $f^{+}_{p}(x) = x^3 \cdot 1 - 4x \cdot 1 = x^3 - 4x$ is an upper bound for $f(x)$ on $(-2,0)$.
• $f^{-}_{p}(x) = x^3 \cdot 0 - 4x \cdot 0 = 0$ is a lower bound for $f(x)$ on $(-2,0)$.
• $f^{+}_{n}(x) = x^3 \cdot 0 - 4x \cdot 0 = 0$ is an upper bound for $f(x)$ on $(-\infty,-2)$.
• $f^{-}_{n}(x) = x^3 \cdot 1 - 4x \cdot 1 = x^3 - 4x$ is a lower bound for $f(x)$ on $(-\infty,-2)$.

• Why are there 4 bounds instead of 2 here? The graph of $x^3-4x$ switches from negative to positive at $x=-2$, so an upper bound on one side becomes a lower bound on the other, and vice versa.

From here we can evaluate the limits of the bounds:

• $\lim\limits_{x \rightarrow -2^{+}} f^{+}_{p}(x) = (-2)^3 - 4(-2) = 0$
• $\lim\limits_{x \rightarrow -2^{+}} f^{-}_{p}(x) = 0$
• $\lim\limits_{x \rightarrow -2^{-}} f^{+}_{n}(x) = 0$
• $\lim\limits_{x \rightarrow -2^{-}} f^{-}_{n}(x) = (-2)^3 - 4(-2) = 0$

The limits from both the right and left side of both bounds converge to $0$. Thus by the Squeeze Theorem, $\lim\limits_{x \rightarrow -2} f(x) = 0$.

5. Let $f(x) = \dfrac{\sin(x)+1}{x}$

1. Find an upper bound and a lower bound for $f(x)$ on $(0,\infty)$.
2. Graph $f(x)$ and both bounds.
3. Use the bounds to find $\lim\limits_{x \rightarrow \infty} f(x)$.
1. Notice that the range of $\sin$ is $[-1,1]$. Thus, an upper bound for $\sin(x)$ is 1, and a lower bound is $-1$. Substituting in these bounds into $f(x)$, we can get an upper bound for $f(x)$ of $f^+(x)$ and a lower bound of $f^-(x)$:

• $f^+(x)=\dfrac{1+1}{x} = \dfrac{2}{x}$
• $f^-(x)=\dfrac{1-1}{x} = 0$
2. 3. Clearly $\lim\limits_{x \rightarrow \infty} \dfrac{1}{x} = 0$ and $\lim\limits_{x \rightarrow \infty} 0 = 0$. Thus by the Squeeze Theorem, $\lim\limits_{x \rightarrow \infty} \dfrac{\sin(x)+1}{x} = 0$.