The function $f(x)=\begin{cases}x^n\sin(1/x), & x\neq0\\\\0,& x=0\end{cases}$ is continuous and differentiable at $x=0$ if
- A. $n\in(0,1]$
- B. $n\in[1,\infty)$
- C. $n\in(1,\infty)$
- D. $n\in(-\infty,0)$
✅ Correct Answer: $n\in(1,\infty)$
Explanation
For continuity, $n>0$. For differentiability, limit $\lim_{x\to0}\frac{f(x)-0}{x}=\lim x^{n-1}\sin(1/x)$ must exist, which happens for $n>1$.
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