Prove $\lim_{x\to 0} f(x)$ exists $\implies\lim_{x\to 0} f(x^2)$ exists Want to show $\lim_{x\to 0} f(x)$ exists $\implies \lim_{x\to 0} f(x^2)$ exists.  (I recently posted a similar variation of the question, where I asked if it was true for $a$ rather than $0$ which can be found here: Prove that $\lim_{x\to0} f(bx)$ exists, if $\lim_{x\to0} f(x)$ exists)  Anyway, here is my attempted proof:
Assume $\lim_{x\to 0} f(x)=l$.  Then $\forall\epsilon_0 \exists\delta_0 \forall x:0<|x|<\delta_0 \implies |f(x)-l|<\epsilon_0$
It follows that $\forall\epsilon_0 \exists\delta_0 \forall x:0<|x^2|<\delta_0 \implies |f(x^2)-l|<\epsilon_0$ (this is the part I was most questionable about - am I allowed to do this?).
Now let $g(x)=x^2$, and it can be shown that $\lim_{x\to 0} x^2 = 0$.  Hence $\forall\epsilon_1 \exists\delta_1 \forall x:0<|x|<\delta_1 \implies |x^2|<\epsilon_1$.  Then let $e_1 = \delta_0.$ And hence from the statement in the last line, there exists a $\delta_1$ for which $|x^2|<\epsilon_1 = \delta_0$
Then for a given $\epsilon$ I take my $\delta = \delta_1$.  Then it follows that $\forall x:0<|x|<\delta \implies |f(x^2)-l|<\epsilon$
 A: 
(this is the part I was most questionable about - am I allowed to do this?).

You are.  But if it makes you uncomfortable skip it and work back.
$\lim_{x\to 0} f(x) = L$ is given. so: For any $\epsilon>0$ there is a $\delta>0$ so that $0<|x| < \delta \implies |f(x) - L| < \epsilon$.
And $\lim_{x\to 0}x^2 = 0$ because for any $\epsilon >0$ there is a $\delta=\sqrt \epsilon> 0$ so that $0<|x|<\delta\implies |x^2| < \delta^2 =\epsilon$.
So for any $\epsilon$ let $\delta_{inbetween}$ be a value so that $0< |x|<\delta_{inbetween} \implies |f(x)-L| < \epsilon$ and let $\delta$ be a value so that $0<|x|<\delta \implies |x^2| < \delta_{inbetween}$.
Then $0< |x|< \delta \implies 0< |x^2| < \delta_{inbetween} \implies |f(x^2)-L| < \epsilon$.
So $\lim_{x\to 0} f(x^2) = L$.
A: It suffices to observe that as $x\to0\implies x^2\to 0$. 
Note that the same property holds also for $x\to1$.
More in general it is true that
$$\lim_{x\to a} f(x) \quad \text{exists}\implies \lim_{x\to a} f(g(x)) \quad \text{exists}$$
for any $g(x)$ such that $g(x)\to a$ with $g(x)\neq a$.
We are excluding the cases for which $f(a)\neq\lim_{x\to a} f(x)$ or $f(a)$ is undefined and $g(x)=a$.
