Sandwich Theorem

Some limits cannot be evaluated by any of the algebra-of-limits theorems directly — typically because the expression involves a factor, such as $\sin(1/x)$, that oscillates and has no limit of its own. The Sandwich Theorem (also called the Squeeze Theorem) sidesteps this by trapping the troublesome function between two better-behaved functions that share a common limit.

Theorem 9.5 (Sandwich Theorem). Suppose $g,f,h:I\to\mathbb R$ satisfy $g(x)\le f(x)\le h(x)$ for every $x$ in a punctured neighbourhood of $x_0$ contained in $I$. If $\lim_{x\to x_0} g(x)=\lim_{x\to x_0} h(x)=l$, then $\lim_{x\to x_0} f(x)=l$ as well.

Intuitively (Fig. 9.27), if $f$ is permanently squeezed between a floor function $g$ and a ceiling function $h$, and floor and ceiling are forced together to the same height $l$ right at $x_0$, then $f$ — having nowhere else to go — is forced to that same height too.

Worked technique (Example 9.29). To evaluate $\lim_{x\to0} x^2\sin\!\left(\frac1x\right)$: since $\sin$ of anything always lies in $[-1,1]$, we have $-x^2\le x^2\sin\!\left(\frac1x\right)\le x^2$ for every $x\ne0$. Both bounding functions $g(x)=-x^2$ and $h(x)=x^2$ tend to $0$ as $x\to0$, so by the Sandwich Theorem the squeezed function also tends to $0$.

A simpler application (Example 9.30). Since $-|\theta|\le\sin\theta\le|\theta|$ for every real $\theta$, and both $-|\theta|$ and $|\theta|$ tend to $0$ as $\theta\to0$, the Sandwich Theorem immediately gives $\lim_{\theta\to0}\sin\theta=0$ — a fact that will itself be used as a stepping stone in the very next section. …