Question related to L'Hopital's Theorem

Question

If $\lim_{x\to0} \frac{f(x)}{g(x)} = L$, what is $\lim_{x\to0} \frac{f(x^2)}{g(x^2)}$?

And when $a$ is not $0$, even if $\lim_{x\to a} \frac{f(x)}{g(x)} = L$, why cannot solve $\lim_{x\to a} \frac{f(x^2)}{g(x^2)}$?

Answer 1

Substitute: $t=x^2$. Then obviously we have that $t \to a^2$ as $x \to a$. This means that:

$$\lim_{x \to a} \frac{f(x^2)}{g(x^2)} = \lim_{t \to a^2} \frac{f(t)}{g(t)}$$

But it's obvious that we don't have any information about the limit, unless $a^2=a \iff a =0,1$

Answer 2

We can say that

$\lim_{x\to a}\frac{f(x^2)}{g(x^2)}= L\implies$

$\lim_{x\to a^2}\frac{f(x)}{g(x)}=L$.

and

if $a>0$ then

$\lim_{x \to a}\frac{f(x)}{g(x)}=L$ $\implies$

$\lim_{x\to \sqrt{a}}\frac{f(x^2)}{g(x^2)}=L$.