The Squeeze Theorem
The sandwich (aka squeeze) theorem
is very useful for computing limits like $\displaystyle \lim_{x \to
0} \frac{\sin(x)}{x}$ that would be difficult to evaluate
otherwishe. It says that if $g(x)$ is sandwiched between
$f(x)$ and $h(x)$, and $f(x)$ and $h(x)$ have the same limit $L$ as
$x \to a$, then $g(x)$ also approaches $L$ as $x\to a$:
Theorem:
Suppose that $f(x) \le g(x) \le h(x)$ for all $x$ that are
close to (but not equal to) $a$, and that
$\displaystyle\lim_{x \to a} f(x) = \lim_{x \to a}
h(x) = L$. Then $\displaystyle\lim_{x \to a} g(x) =
L$. 
When $x$ is close to $a$, $f(x)$ and $h(x)$ are
close to $L$. So $g(x)$ is somewhere between (a number close to
$L$)
and (another number close to $L$). This means that $g(x)$
must be
close to $L$!
Example: Evaluate $\displaystyle{\lim_{x \to 0}
x^2 \sin\left(1/x\right)}$.
DO: Why is it true that
$1\le \sin\theta\le 1$ no matter what $\theta$ is?
Solution: Since $1 \le \sin(1/x)\le 1$, we can multiply
all three sides by the ($\ge 0$) value $x^2$, getting $$x^2 \le
x^2 \sin(1/x)
\le x^2.$$
so $g(x)=x^2 \sin(1/x)$ is squeezed
(or sandwiched)
between $f(x)=x^2$ and $h(x)=x^2$. Since both $x^2$ and $x^2$
approach 0 as $x \to 0$, we must also have $$\lim_{x \to 0} x^2
\sin(1/x) = 0.$$
