Miscellaneous Question on Limits

here, lim (f(x^2)-f(x))/(f(x)-f(0)) = lim ((f(x^2)-f(x))/x)/((f(x)-f(0))/x) Now the denominator is f'(0). so lets consider the numerator-: lim (f(x^2)-f(x))/x = lim (f(x^2)-f(0))/x - (f(x)-f(0))/x = (fog)'(0) - f'(0) where g(x) = x^2. hence the numerator becomes 2(0)f'(0) - f'(0) = -f'(0) and hence the final answer is -f'(0)/f'(0) = -1

Due to f(x) is differentiable, $f(x)$ ~ $f(0) + x \cdot f '(0)$ when x $\rightarrow$ 0. ( $f(x) = f(0) + x \cdot f '(0) + o(|x|)$ when x $\rightarrow$ 0)

$f(x^2)$ ~ $f(0) + x^2 \cdot f '(0)$ when x $\rightarrow$ 0. Therefore, $\lim_{x \rightarrow 0} \frac{f(x^2) - f(x)}{f(x) - f(0)} =$ $= \lim_{x \rightarrow 0} \frac{(x^2 - x)f '(0)}{xf '(0)}$ $= \lim_{x \rightarrow 0} \frac{(x(x -1))f '(0)}{xf '(0)} = \lim_{x \rightarrow 0} x - 1 = \boxed{-1}$

Note: I haven't used L'Hopital Rule nor that f(x) is a strictly increasing function