Squeeze Theorem Question My question is from this Video
In the last example
He says that 
$$\lim_{x \to 0} x^2 \cos(\frac{1}{x^2}) = 0$$
Squeeze Theorem:
$$g(x) \leq f(x) \leq h(x)$$
Given:
$$-1 \leq \cos(x) \leq 1$$
he confusing gets 
$$-x^2 \leq x^2\cos(\frac{1}{x^2})\leq x^2$$
and finds the limits with that.
How does he find $g(x) \text{ and } h(x)$ for the squeeze theorem. Is there a special way to find them?
 A: Let $ a \leq y \leq b$ be an inequality for $a,b \in \mathbb{R}$. If we multiply this inequality by a positive number $c$, this does not change the sense of the inequalty. Thus,
$$  a \leq y \leq b \iff   ac \leq cy \leq cb, c >0 .$$
Hence, take $a = -1, b=1$, $c= x^2$ and $y = \text{cos}(\frac{1}{x^2}),$ and the result follows*.
*The goal is to find functions $g(x)$ and $h(x)$ that have the same limit, such that $g(x) \leq f(x) \leq h(x)$.
A: I don't know whether there is any special way to find $g(x)$ and $h(x).$ It's mostly intuition and using some known inequality. Here are some examples:
(1). $\lim_{x \to 0} \sin x = 0.$ In this case we use the fact that $|\sin x | \leq 1.$ So we get $-x \leq \sin x \leq x.$
(2). $\lim_{x \to 0} \cos x = 1.$ In this case we use  the following inequality $1 - \frac{1}{2}x^2 \leq \cos x \leq 1, \forall x \in \mathbb R.$
(3). $\lim_{x \to 0}\dfrac{\cos x -1}{x} = 0.$ For this we use two inequality:
$$
-\frac{1}{2}x \leq \dfrac{\cos x -1}{x} \leq x, \text{for} \space x > 0
$$
and
$$
0 \leq \dfrac{\cos x -1}{x} \leq -\frac{1}{2}x, \text{for} \space x < 0.
$$
Now define $g(x) := -\frac{x}{2}$ for $x \geq 0$ and $g(x) := 0$ for $x < 0.$ Also define $h(x) := 0$ for $x \geq 0$ and $h(x) := -\frac{x}{2}$ for $x < 0.$
If you look at the above examples, then you can see that, to use squeeze theorem, we are using some known inequality so that both $g(x)$ and $h(x)$ have the same limit. The more complicated the given functions is, the harder it is to find $g(x), h(x)$ to use squeeze theorem.
