Question
If the function $f(x)=\left\{\begin{array}{cc}\frac{\sin x^2}{x} ; & x \neq 0 \\ 0 ; & x=0\end{array}\right.$, is differentiable at $x=0$, then right hand derivative of $f(x)$ at $x=0$ is
✓
Answer
(c) : At $x=0$, right hand derivative
$
\begin{aligned}
f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\frac{\sin h^2}{h}-0}{h}=\lim _{h \rightarrow 0} \frac{\sin h^2}{h^2}=1
\end{aligned}
$
$
\begin{aligned}
f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\frac{\sin h^2}{h}-0}{h}=\lim _{h \rightarrow 0} \frac{\sin h^2}{h^2}=1
\end{aligned}
$
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