Prove $\lim_{x\to a} f(x)$ exists $\implies \lim_{x\to a} f(x^2)$ exists

Question

Want to prove rigorously (if possible, since I was not able to think of any counter-example) that $\lim_{x\to a} f(x)$ exists $\implies \lim_{x\to a} f(x^2)$ exists. (I also have a feeling that the limits equate.)

I started with the $\epsilon-\delta$ definition of $\lim_{x\to a} f(x) = l$ which states that $\forall \epsilon >0 \exists \delta>0 \forall x:0<|x-a|<\delta \implies |f(x)-l| < \epsilon$.

Now I want to show that $\forall \epsilon >0 \exists \delta>0 \forall x:0<|x-a|<\delta \implies |f(x^2)-l_1| < \epsilon$ where $l_1 \in \mathbb{R}$.

I've got no clue how to continue. All help is appreciated.

As an additional note: I believe the converse ($\lim_{x\to a} f(x^2)$ exists $\implies \lim_{x\to a} f(x)$ exists) is not true as $f(x)=\frac{|x|}{x}$ is a counter-example.

Edit: A counter-example was found for the statement but I was wondering whether it is true for $a = 0$? So in other words, I would like to prove or disprove the statement $\lim_{x\to 0} f(x)$ exists $\implies \lim_{x\to 0} f(x^2)$ exists

Answer 1

This is false, as evidenced by the choices $a=2$ and $f(x) = \begin{cases} x, & x \leq 4 \\ x+1, & x>4.\end{cases}$

Answer 2

It is always true for $a=0$ and $a=1$ otherwise just assume a discontinuous function at $x=a^2$.