The specific problem is: $\lim\limits_{x \to -1^-} f(x^2)$
My so far attempting a substitution method: $$x\to -1^-$$ $$u = x^2$$ $$u\to 1^\text{not sure what this is now}$$
I can't really use the method to solve this, yet, since my underlying is unresolved: how does the $x^2$ affect the left direction of the limit?
Additionally, is there a better to solve this? After evaluating the $u\to 1^c$ (c is either + or - aka left or ) I was planning to then substitute like this:
$$\lim_{u \to 1^c} f(u)$$
$x\to a^-$ means, that $x$ is approaching $a$ on the valueas that are smaller than $a$.
So if $x\to y=a-\epsilon$ for $\epsilon \to 0^+$ ($\epsilon \geq 0$), then
$x^2 \to y^2 = (a-\epsilon)^2 = a^2 -2 a\epsilon +\epsilon^2$.
Since $a\leq 0$, we have
$-2 a\epsilon\geq 0$,
thus
$a^2 -2 a\epsilon +\epsilon^2 \geq a^2$
so for $a\leq 0$ we have $$x^2 \to (a^2)^+$$
EDIT
I've been thinking about simplier eplaination.
If $i\leq j\leq0$, then $0\leq j^2 \leq i^2$.
In our case we have $x\leq -1$, so $1\leq x^2$. Thus $x^2\to 1^+$