Problem : Let f(x) be strictly increasing and differentiable then $\lim_{x\to 0} \frac{f(x^2) -f(x)}{f(x) -f(0)}$ is..... Problem : Let f(x) be strictly increasing and differentiable then $\lim_{x\to 0} \frac{f(x^2) -f(x)}{f(x) -f(0)}$ is.....
My approach : 
Sol. Using L Hospital's rule $= \lim_{x \to 0} \frac{2xf'(x^2)-f'(x)}{f'(x)}$ 
$ = -1 +\lim_{x\to 0} \frac{2xf'(x^2)}{f'(x)}$
Now how to proceed further I refer the solution it is given since $f'(x) >0$ the answer is -1, but didn't get that, please guide on this will be of great help. 
 A: The question has unnecessary assumptions. What we need here is the fact that $f'(0) \neq 0$. Under this assumption we can see that
\begin{align}
L &= \lim_{x \to 0}\frac{f(x^{2}) - f(x)}{f(x) - f(0)}\notag\\
&= \lim_{x \to 0}\frac{f(x^{2}) - f(0) - \{f(x) - f(0)\}}{f(x) - f(0)}\notag\\
&= \lim_{x \to 0}\frac{f(x^{2}) - f(0)}{f(x) - f(0)} - 1\notag\\
&= \lim_{x \to 0}x\cdot\frac{f(x^{2}) - f(0)}{x^{2}}\cdot\frac{x}{f(x) - f(0)} - 1\notag\\
&= 0 \cdot f'(0)\cdot\frac{1}{f'(0)} - 1 = -1\notag
\end{align}
If $f'(0) = 0$ then it is not possible to conclude whether this limit exists or not.
Moreover L'Hospital's Rule is not useful here because we don't have enough information about behavior of $f'(x)$.
A: You can use the logic that:
Since $f$ is strictly increasing and differentiable, $f'(x)>0$.
So $$\lim_{x\to 0} \frac{2xf'(x^2)}{f'(x)}= \frac{2\cdot 0 f'(0)}{f'(0)}=0$$ since we have already remarked that $f'(0) \not = 0$ due to the aforementioned reason.
And your final limit is $-1$ as the answer says.
A: suppose $f'(0)$ exists, then you have $$\begin{align}f(x) &=  f(0) + x f'(0) + \frac12 x^2 f''(0) + \cdots\\
f(x^2) &= f(0) + x^2f'(0)+ \cdots\\
\frac{f(x^2) - f(x)}{f(x) - f(0)} &= \frac{-xf'(0)-x^2\left(f'(0)-\frac12 f''(0)+\cdots\right)}{xf'(0)+\cdots} = -1+\cdots\end{align}$$  therefore $$\lim_{x \to 0}\frac{f(x^2) - f(x)}{f(x) - f(0)} = -1. $$
