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Q.
The set of points, where $f\left(x\right)=\frac{x}{1+\left|x\right|} is$
differentiable, is :
AIEEEAIEEE 2008
Solution:
$\because f \left(x\right)=\frac{x}{1+\left|x\right|}$
Let $f \left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}=\frac{x}{1+\left|x\right|}$
It is clear that $g (x) = x$ and $h (x) = 1 + |x |$ are differentiable on $\left(-\infty, \infty\right)$ and $\left(-\infty, 0\right)\cup \left(0, \infty\right)$ respectively.
Thus $f (x)$ is differentiable on $ \left(-\infty, 0\right)\cup \left(0, \infty\right).$ For $x = 0$
$\displaystyle \lim_{x \to 0}$$\frac{f\,\left(x\right)-f\,\left(0\right)}{x-0} =$ $\displaystyle \lim_{x \to 0}$$\frac{\frac{x}{1+\left|x\right|}}{x}$
$\displaystyle \lim_{x \to 0}$$\frac{1}{1+\left|x\right|}=1$
Thus $f (x)$ is differentiable on $\left(-\infty, \infty\right).$