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(D) Given the function $f(x) = \begin{cases} \frac{\sin(x^2)}{x} & x 
eq 0 \ 0 & x = 0 \end{cases}$.First,check for continuity at $x=0$:$\lim_{x i

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differentiable and the derivative is continuous

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(D) Given the function $f(x) = \begin{cases} \frac{\sin(x^2)}{x} & x 
eq 0 \ 0 & x = 0 \end{cases}$.First,check for continuity at $x=0$:$\lim_{x ightarrow 0} f(x) = \lim_{x ightarrow 0} \frac{\sin(x^2)}{x} = \lim_{x ightarrow 0} x \cdot \frac{\sin(x^2)}{x^2} = 0 \cdot 1 = 0 = f(0)$.Since the limit equals the function value,$f(x)$ is continuous at $x=0$.Next,check for differentiability at $x=0$:$f'(0) = \lim_{h ightarrow 0} \frac{f(0+h) - f(0)}{h} = \lim_{h ightarrow 0} \frac{\frac{\sin(h^2)}{h} - 0}{h} = \lim_{h ightarrow 0} \frac{\sin(h^2)}{h^2} = 1$.Since the limit exists,$f(x)$ is differentiable at $x=0$ and $f'(0) = 1$.Now,find $f'(x)$ for $x 
eq 0$:$f'(x) = \frac{d}{dx} \left( \frac{\sin(x^2)}{x} ight) = \frac{x \cdot \cos(x^2) \cdot 2x - \sin(x^2) \cdot 1}{x^2} = 2\cos(x^2) - \frac{\sin(x^2)}{x^2}$.Check if $f'(x)$ is continuous at $x=0$:$\lim_{x ightarrow 0} f'(x) = \lim_{x ightarrow 0} \left( 2\cos(x^2) - \frac{\sin(x^2)}{x^2} ight) = 2(1) - 1 = 1$.Since $\lim_{x ightarrow 0} f'(x) = f'(0) = 1$,the derivative is continuous at $x=0$.Thus,$f$ is differentiable and the derivative is continuous.

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} \frac{\sin(x^2)}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$. Then,at $x=0$,$f$ is

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