Question

Let $f(x)=\begin{cases}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{cases}$Then at $x=0$

• A. $f$ is continuous but not differentiable
• B. $f$ is continuous but $f^{\prime}$ is not continuous
• C. $f^{\prime}$ is continuous but not differentiable
• D. $f$ and $f^{\prime}$ both are continuous

Answer 1

The Correct Option is B

To determine the continuity and differentiability of the function \( f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x = 0 \end{cases} \) at \( x = 0 \), let us analyze it step-by-step. 

1. Check if the function \( f(x) \) is continuous at \( x = 0 \):
   • The function is continuous at a point if the following conditions are satisfied:
     1. The function is defined at the point.
     2. The limit of the function as it approaches the point exists.
     3. The limit of the function as it approaches the point is equal to the function's value at that point.
   • Here, \( f(0) = 0 \).
   • Calculate \( \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \).
   • We know \(\left| \sin\left(\frac{1}{x}\right) \right| \leq 1\), therefore: \(-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2\).
   • As \( x \to 0 \), both \(-x^2\) and \(x^2\) approach 0. By the Squeeze Theorem, \(\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0\).
   • Therefore, \(\lim_{x \to 0} f(x) = 0 = f(0)\), meaning \( f \) is continuous at \( x = 0 \).
2. Check if the derivative \( f^{\prime}(x) \) is continuous at \( x = 0 \):
   • To find \( f^{\prime}(x) \), we'll use the definition of derivative: \[ f^{\prime}(x) = \text{if } x \neq 0, \text{ using the product rule: } \] \[ f^{\prime}(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \]
   • For \( x = 0 \), the derivative by definition is: \[ f^{\prime}(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0 \] by the squeeze theorem again.
   • We need to check the continuity of \( f^{\prime}(x) \) at \( x = 0 \): \[ \lim_{x \to 0} f^{\prime}(x) = \lim_{x \to 0} \left( 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right) = \text{does not exist} \] due to the oscillating behavior of \(-\cos\left(\frac{1}{x}\right)\).
   • Since \(\lim_{x \to 0} f^{\prime}(x)\) does not exist, \( f^{\prime}(x) \) is not continuous at \( x = 0 \).

In conclusion, \( f \) is continuous at \( x = 0 \), but \( f^{\prime} \) is not continuous at \( x = 0 \). Therefore, the correct answer is: $f$ is continuous but $f^{\prime}$ is not continuous.