Suppose that $I$ is an interval, and $g,f,h$ are functions defined on $I$, except possibly at $x_0$. Furthermore, $\forall \varepsilon>0,$ $\forall x\in I,$ and for some constant $L$,
$$ (L-\varepsilon)g(x) < f(x) \le Lh(x) .$$
Also,
$$\lim_{x\to x_0} g(x) = \lim_{x\to x_0} h(x) = 1.$$
Then is it true that $\forall \varepsilon>0,$
$$L-\varepsilon < \lim_{x\to x_0} f(x) \le L,$$
and since $\varepsilon$ can be taken to be arbitrarily small, the middle is forced to give
$$\lim_{x\to x_0} f(x)=L?$$
Note that this is essentially the Squeeze Theorem with a strict lower-bound.
[Math] Squeeze Theorem with strict inequalities
analysiscalculusreal-analysis
Related Solutions
Let $\varepsilon > 0$ be given. By the convergence of the series $\sum a_n$ resp. $\sum c_n$, there is an $N(\varepsilon)$ such that for all $N(\varepsilon) \leqslant k < m$ we have
$$\left\lvert\sum_{n=k+1}^m a_n \right\rvert < \varepsilon\land \left\lvert\sum_{n=k+1}^m c_n \right\rvert < \varepsilon.\tag{1}$$
By the inequalities $a_n \leqslant b_n \leqslant c_n$, we have
$$\sum_{n=k+1}^m a_n \leqslant \sum_{n=k+1}^m b_n \leqslant \sum_{n=k+1}^m c_n.$$
Now $(1)$ implies
$$-\varepsilon < \sum_{n=k+1}^m b_n < \varepsilon$$
for $N(\varepsilon) \leqslant k < m$.
Since $\varepsilon$ was arbitrary, the sequence of partial sums is a Cauchy sequence, and $\sum b_n$ converges.
Squeeze theorem can only be applied where there is a squeeze, i.e. the function is sandwiched between two other functions, and the bread functions have the same value at the point where you need to calculate the limit.
As in your case, both $x^2$ and $-x^2$ approach zero. The squeeze theorem looks beautiful graphically. I am posting some examples
$$x^{2}\ge x^{2}\sin\left(\frac{1}{x}\right)\ge-x^{2}$$
$$x\ge x\sin{x}\ge-x$$
Squeeze theorem, in general, is effective for calculating limits of product of functions. It also proves useful while finding limit of infinite sums.
But for the cases where you can't use squeeze theorem, other techniques have been devised. It isn't necessary that one should expect a general approach to every problem in a vast topic like limits.
Best Answer
Suppose that $L$ is fixed. The quantified statement $$(\forall \epsilon > 0) (\forall x \in I) \quad (L-\epsilon)g(x) < f(x) \le L h(x)$$ implies the quantified statement $$(\forall x \in I) \quad L g(x) \le f(x) \le L h(x).$$ Now apply the squeeze theorem.