Q. 81 - Calculus

Question

Use the Squeeze Theorem to find the limits. Explain exactly how the Squeeze Theorem applies in each case.

$\underset{x\to 0}{\lim }\left(e^x-1\right) \sin \frac{1}{x}$

Step-by-Step Solution

Answer

The limit of the given equation is $0$.

Step 1. Given information.

Consider the given question,
$\underset{x\to 0}{\lim }\left(e^x-1\right) \sin \frac{1}{x}$

Step 2. Apply squeeze theorem.

Range of $\sin \frac{1}{x}$ or $\sin x$ is $\left[-1,1\right]$.

$-1\le \sin \frac{1}{x}\le 1$

Multiply all the sides by $\left(e^x-1\right)$,

$$\begin{gathered} -1\left(e^x-1\right)\le \sin \frac{1}{x}\left(e^x-1\right)\le 1\left(e^x-1\right) \\ -e^x+1\le \sin \frac{1}{x}\left(e^x-1\right)\le e^x-1 \end{gathered}$$

Applying limits on all the sides as $x\to 0$,

$$\begin{gathered} \underset{x\to 0}{\lim }\left(-e^x+1\right)\le \underset{x\to 0}{\lim }\sin \frac{1}{x}\left(e^x-1\right)\le \underset{x\to 0}{\lim }\left(e^x-1\right) \\ -e^0+1\le \underset{x\to 0}{\lim }\sin \frac{1}{x}\left(e^x-1\right)\le \underset{x\to 0}{\lim }\le e^0-1 \\ -1+1\le \underset{x\to 0}{\lim }\sin \frac{1}{x}\left(e^x-1\right)\le 1-1 \\ 0\le \underset{x\to 0}{\lim }\sin \frac{1}{x}\left(e^x-1\right)\le 0 \end{gathered}$$

Applying the Squeeze theorem,

$=0$

Step 3. Plot the graph.

Representing the graph, we get,