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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