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