There's no general way of finding two functions that satisfy the criteria...
You just want to find two functions g and h such that:
lim_{x\rightarrow a}g(x)=lim_{x\rightarrow a}h(x) and
g(x)\leq f(x) \leq h(x) for all x within ssome neighbourhood of a. Then, the squeeze theorem tells you that the limit those two have at a is the same as the limit f has at a.
Like I said-- the choice of your g and h is completely arbitrary-- you just want them to satisfy those conditions.
Sometimes, e.g. when you have \frac{1}{x}sin(x), boundedness helps. Observe:
-1 \leq sin(x) \leq 1 (Property of the sine function)
This implies -\frac{1}{x} \leq \frac{1}{x} sin(x) \leq \frac{1}{x}. (Whenever x>0)
As you can see, the limit as x\rightarrow \infty of the left and right hand sides of the inequality match and equal 0, and so the limit of the function \frac{1}{x} sin(x) is 0. This is a classic application of the squeeze theorem.
I hope that helped.