My textbook (Calculus Early Transcendentals by James Stewart) doesn't provide an explanation or general method for finding functions that bound the function whose limit is being taken. It only provides the example $$\lim_{x \to 0} x^2 \sin\left(\frac {1}{x}\right)$$
And says that sine is always between $-1$ and $1$ so essentially the upper and lower output values of sine are the functions which bound sine.
$$-1 \leq \sin\left(\frac {1}{x}\right) \leq 1$$
Then multiplying the inequality by $x^2$ gives
$$-x^2 \leq x^2\sin\left(\frac {1}{x}\right) \leq x^2$$
Where now it is evident that $-x^2$ and $x^2$ are the functions which bound $x^2 \sin\left(\frac {1}{x}\right)$.
But for limits where this method of first using the upper and lower output values of sine or cosine to establish an inequality doesn't work, how do I find the functions which bound the function whose limit is being taken?
In other words, what are some strategies for finding "bounding" functions? Or how does one think through such a problem of finding "bounding" functions?
Best Answer
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.