Question

Let $f\left(x\right)=\frac{\sin x^2}{x},x\ne 0,andf\left(x\right)=0$, at $x=0$ then $f^{\prime }\left(0^{+}\right)+f^{\prime }\left(0^{-}\right)$ has the value equal to

Answer 1

$f\left(x\right)=\frac{\sin x^2}{x},x\ne 0$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin h^2}{h}-0}{h}$
$=\underset{h\to 0}{lim}\frac{\sin h^2}{h^2}$
$f^{\prime }\left(0^{+}\right)=1$
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin (-h{)}^2}{-h}-0}{-h}$
$=\underset{h\to 0}{lim}\frac{\sin h^2}{h^2}$
$f^{\prime }\left(0^{-}\right)=1$

So, $f^{\prime }\left(0^{+}\right)+f^{\prime }\left(0^{-}\right)=1+1=2$