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 $$\lim_{x \to 0} x^2 \sin(1/x) = 0.$$