Q. 80 - Calculus

Question

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

$\underset{x\to 0}{\lim }x \sin \frac{1}{x^2}$

Step-by-Step Solution

Answer

The limit of the given equation is $0$.

Step 1. Given information.

Consider the given question,

$\underset{x\to 0}{\lim }x \sin \frac{1}{x^2}$

Step 2. Apply squeeze theorem.

Range of $\sin \frac{1}{x^2}$ or $\sin x^2$ is $\left[-1,1\right]$.

$-1\le \sin \frac{1}{x^2}\le 1$

Multiply x on all the sides,

$$\begin{gathered} -1\left(x\right)\le \left(x\right)\sin \frac{1}{x^2}\le 1\left(x\right) \\ -x\le x \sin \frac{1}{x^2}\le x \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 0}{\lim }\left(-x\right)\le \underset{x\to 0}{\lim } x \sin \frac{1}{x^2}\le \underset{x\to 0}{\lim }\left(x\right) \\ -0\le \underset{x\to 0}{\lim } x^2 \sin \frac{1}{x^2}\le 0 \end{gathered}$$

Applying the Squeeze theorem,

$=0$

Step 3. Plot the graph.

Representing the graph, we get,