Chapter 2, Problem 4RCC

To determine

To Describe: The Squeeze Theorem

Explanation of Solution

Statement:

Suppose that $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$  for all x in some open interval containing c except possibly at c itself. If $\underset{x\to c}{\lim }g\left(x\right)=L=\underset{x\to c}{\lim }h\left(x\right)$ then $\underset{x\to c}{\lim }f\left(x\right)=L$.

Result used: Definition of limits

Let $\epsilon >0$ for any $\delta >0$ such that $\left|f\left(x\right)-L\right|<\epsilon$ whenever $0<\left|x-c\right|<\delta$.

By the definition of limits, since $\underset{x\to c}{\lim }g\left(x\right)=L$ there exist some ${\delta }_1>0$ such that

$\left|g\left(x\right)-L\right|<\epsilon$ for all $0<\left|x-c\right|<{\delta }_1$

Proof:

Let $\epsilon >0$ for any $\delta >0$ such that $\left|f\left(x\right)-L\right|<\epsilon$ whenever $0<\left|x-c\right|<\delta$.

By the definition of limits, since $\underset{x\to c}{\lim }g\left(x\right)=L$ there exist some ${\delta }_1>0$ such that

$\left|g\left(x\right)-L\right|<\epsilon$ for all $0<\left|x-c\right|<{\delta }_1$.

Thus, $-\epsilon <g\left(x\right)-L<\epsilon$ for all $0<\left|x-c\right|<{\delta }_1$

$L-\epsilon <L+g\left(x\right)-L<L+\epsilon$ for all $0<\left|x-c\right|<{\delta }_1$

$L-\epsilon <g\left(x\right)<L+\epsilon$ for all $0<\left|x-c\right|<{\delta }_1$ (1)

Similarly, since $\underset{x\to c}{\lim }h\left(x\right)=L$ there exist some ${\delta }_2>0$ such that

$\left|h\left(x\right)-L\right|<\epsilon$ for all $0<\left|x-c\right|<{\delta }_2$.

Thus, $-\epsilon <h\left(x\right)-L<\epsilon$ for all $0<\left|x-c\right|<{\delta }_2$

$L-\epsilon <L+h\left(x\right)-L<L+\epsilon$ for all $0<\left|x-c\right|<{\delta }_2$

$L-\epsilon <h\left(x\right)<L+\epsilon$ for all $0<\left|x-c\right|<{\delta }_2$ (2)

Since $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$ for all x in some open interval containing c, there exist ${\delta }_3>0$ such that $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$ for all $0<\left|x-c\right|<{\delta }_3$ (3)

Choose $\delta =$ min $\left({\delta }_1,{\delta }_2,{\delta }_3\right)$ then by (1),(2) and (3) equations

$L-\epsilon <g\left(x\right)\le f\left(x\right)\le h\left(x\right)<L+\epsilon$ for all $0<\left|x-c\right|<\delta$

$L-\epsilon <f\left(x\right)<L+\epsilon$ for all $0<\left|x-c\right|<\delta$

$L-\epsilon -L<f\left(x\right)-L<L+\epsilon -L$ for all $0<\left|x-c\right|<\delta$

Therefore, $-\epsilon <f\left(x\right)-L<\epsilon$ for all $0<\left|x-c\right|<\delta$,

So $\left|f\left(x\right)-L\right|<\epsilon$ for all $0<\left|x-c\right|<\delta$

By the definition of limits $\underset{\_}{\underset{x\to c}{\lim }f\left(x\right)=L}$.

Example:$\underset{x\to \infty }{\lim }\frac{\sin x}{x}=0$

To prove: The limit of the function $\underset{x\to \infty }{\lim }\frac{\sin \left(x\right)}{x}=0$.

Theorem used: The Squeeze Theorem.

“If $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ when xis near a (except possibly at a) and $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }h\left(x\right)=L$ then $\underset{x\to a}{\lim }g\left(x\right)=L$.”

Proof:

First take that $-1\le \sin x\le 1$, because of the well-known properties of the sine function.

Since computing the limit as x goes to infinity, then assume that $x>0$.

Thus, $-\frac{1}{x}\le \frac{\sin \left(x\right)}{x}\le \frac{1}{x}$

Apply the limit that is $\underset{x\to \infty }{\lim }\left(-\frac{1}{x}\right)\le \underset{x\to \infty }{\lim }\left(\frac{\sin \left(x\right)}{x}\right)\le \underset{x\to \infty }{\lim }\left(\frac{1}{x}\right)$

Since $\underset{x\to \infty }{\lim }\left(-\frac{1}{x}\right)=0$ and $\underset{x\to \infty }{\lim }\left(\frac{1}{x}\right)=0$ then $\underset{\_}{\underset{x\to \infty }{\lim }\left(\frac{\sin \left(x\right)}{x}\right)=0}$.