Question

The function $f(x)$ is given by:

$$f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & \text{for } x \neq 0 \\ 0, & \text{for } x = 0 \end{cases}$$

• A. continuous as well as differentiable
• B. differentiable but not continuous
• C. continuous but not differentiable
• D. neither continuous nor differentiable

Hint:

To check continuity at a point, ensure the limit from both sides matches the function value. For differentiability, check the limit of the difference quotient.

Answer 1

The Correct Option is C

The given function is:

$$f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & \text{for } x \neq 0 \\ 0, & \text{for } x = 0 \end{cases}$$

Step 1: Checking continuity at $x = 0$.

For $f(x)$ to be continuous at $x = 0$, we need to check if:

$$\lim_{x \to 0} f(x) = f(0) = 0.$$

For $x \neq 0$, we have:

$$\lim_{x \to 0} x \sin \left( \frac{1}{x} \right).$$

Since $\sin \left( \frac{1}{x} \right)$ is bounded between -1 and 1, we get:

$$- x \leq x \sin \left( \frac{1}{x} \right) \leq x.$$

As $x \to 0$, both bounds approach 0. By the squeeze theorem, we conclude:

$$\lim_{x \to 0} f(x) = 0 = f(0),$$

so the function is continuous at $x = 0$.

Step 2: Checking differentiability at $x = 0$.

The function $f(x)$ is differentiable at $x = 0$ if:

$$\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{x \sin \left( \frac{1}{x} \right)}{x}.$$

This simplifies to:

$$\lim_{x \to 0} \sin \left( \frac{1}{x} \right).$$

Since $\sin \left( \frac{1}{x} \right)$ oscillates infinitely as $x \to 0$, the limit does not exist. Therefore, the function is not differentiable at $x = 0$.

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