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If $f(x)=\frac{x}{1+|x|}$, for $x \in R$, then $f^{\prime}(0)$ is equal to :
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1
We have,
$f(x)=\frac{x}{1+|x|}$
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{h}{1+|h|}-0}{h}$
$=\lim _{h \rightarrow 0} \frac{1}{1+|h|}=1$
$f(x)=\frac{x}{1+|x|}$
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{h}{1+|h|}-0}{h}$
$=\lim _{h \rightarrow 0} \frac{1}{1+|h|}=1$
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