If f(x) is differentiable and strictly increasing function, then the value of $\lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}$, is

-1

We have,

$\lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}=\lim\limits_{x \rightarrow 0} \frac{\left\{f(x)^2-f(0)\right\}-\{f(x)-f(0)\}}{f(x)-f(0)}$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}=\lim\limits_{x \rightarrow 0}\left\{\frac{f\left(x^2\right)-f(0)}{f(x)-f(0)}-1\right\}$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}=\lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(0)}{x^2-0} \times \frac{x^2-0}{f(x)-f(0)}-1$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}=\lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(0)}{x^2} \times \frac{\frac{f(x)-f(0)}{x-0}-1}{x-0}$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}=f'(0) \times \frac{0}{f'(0)}-1=-1~~~~~\left[\begin{array}{r}∵ f'(x)>0 \\ \text { for all } x\end{array}\right]$

Alter

We have,

$\lim\limits_{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}$

$=\lim\limits_{x \rightarrow 0} \frac{2 x f'\left(x^2\right)-f'(x)}{f'(x)}$           [Using De 'L' Hospital's Rule]

$=\lim\limits_{x \rightarrow 0}\left\{\frac{2 x f'\left(x^2\right)}{f'(x)}-1\right\}$

$=2 \times 0 \times \frac{f'(0)}{f'(0)}-1=-1$                   [∵ f'(x) > 0  for all  x]