Q. 82 - Calculus

Question

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

$\underset{x\to 0}{\lim } \sin x \sin \frac{1}{x}$

Step-by-Step Solution

Answer

The limit of the given equation is $0$.

Step 1. Given information.

Consider the given question,

$\underset{x\to 0}{\lim } \sin x \sin \frac{1}{x}$

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 every sides by $\sin x$,

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

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

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

Applying the Squeeze theorem,

$=0$

Step 3. Plot the graph.

Representing the graph, we get,