Differentiability of an Oscillatory Function

Detailed Solution

**Step 1: Continuity at the origin** For $x\ne0$, we have $$ |f(x)|=x^2\left|\sin\left(\frac1x\right)\right|\le x^2, $$ because $|\sin t|\le1$ for every $t\in\mathbb R$. Since $$ \lim_{x\to0}x^2=0, $$ the squeeze theorem gives $$ \lim_{x\to0}f(x)=0=f(0). $$ Therefore, $f$ is continuous at $0$. **Step 2: Differentiability at the origin** Using the definition of the derivative, $$ f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}. $$ For $h\ne0$, $$ \frac{f(h)-f(0)}{h} =h\sin\left(\frac1h\right). $$ Consequently, $$ \left|h\sin\left(\frac1h\right)\right|\le |h|\to0. $$ Hence, $$ \boxed{f'(0)=0}. $$ Thus, $f$ is differentiable at the origin. **Step 3: Derivative for $x\ne0$** Applying the product rule and chain rule, $$ \begin{aligned} f'(x) &=2x\sin\left(\frac1x\right) +x^2\cos\left(\frac1x\right)\left(-\frac1{x^2}\right)\\ &=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right). \end{aligned} $$ Therefore, $$ f'(x)= \begin{cases} 2x\sin\left(\dfrac1x\right)-\cos\left(\dfrac1x\right),&x\ne0,\\[6pt] 0,&x=0. \end{cases} $$ **Step 4: Discontinuity of the derivative at $0$** For continuity of $f'$ at $0$, it is necessary that $$ \lim_{x\to0}f'(x)=f'(0)=0. $$ Consider the sequence $$ x_n=\frac1{2n\pi},\qquad n\ge1. $$ Then $x_n\to0$ and $$ \sin\left(\frac1{x_n}\right)=\sin(2n\pi)=0, $$ while $$ \cos\left(\frac1{x_n}\right)=\cos(2n\pi)=1. $$ Thus, $$ f'(x_n)=-1. $$ Now consider $$ y_n=\frac1{(2n+1)\pi},\qquad n\ge1. $$ Then $y_n\to0$ and $$ \sin\left(\frac1{y_n}\right)=0, \qquad \cos\left(\frac1{y_n}\right)=-1. $$ Therefore, $$ f'(y_n)=1. $$ The two sequences approach $0$, but their derivative values approach different limits: $$ f'(x_n)\to-1,\qquad f'(y_n)\to1. $$ Hence $\lim_{x\to0}f'(x)$ does not exist. In particular, $f'$ is not continuous at $0$. **Verification of the options** - A is false because $f'$ is discontinuous at $0$. - B is true because $f$ is differentiable everywhere, but its derivative is not continuous at $0$. - C is false because $f'(0)=0$ exists. - D is false because $f'(0)=0$, not $1$. Therefore, the correct option is $$ \boxed{\mathrm{B}}. $$