The sandwich theorem.

Theorem 4.6.1   Let $ f$ , $ g$ and $ h$ be three functions defined on the same neighborhood $ \mathcal{V}$ of some real number $ x_0$ (resp. of $ -\infty $ , resp. of $ +\infty$ ). Suppose that:
(i)
For any $ x \in \mathcal{V}$ , the double inequality $ f(x) \leq g(x) \leq h(x)$ holds.
(ii)
The functions $ f$ and $ h$ have the same limit $ L$ at $ x_0$ (resp. at $ -\infty $ , resp. at $ +\infty$ ).
Then $ g$ has a limit at $ x_0$ (resp. at $ -\infty $ , resp. at $ +\infty$ ) and it is equal to $ L$ .

Figure 9: The sandwich theorem for functions.
\begin{figure}\mbox{
\subfigure[]{ \epsfig{file=Sandwich-functions.eps,height=7cm}}
\qquad
\subfigure[]{\epsfig{file=xTimesSin1_x.eps,height=7cm}}
}\end{figure}

Example 4.6.2   Take $ f(x)=x^2$ , $ g(x)=x^2+\sin x / x$ and $ h(x)=x^2+1/x$ . For any real number in $ (0,+\infty)$ , we have $ f(x) \leq g(x) \leq h(x)$ .

Moreover, $ f$ and $ h$ have both a positive infinite limit at $ +\infty$ , thus $ g$ has a positive infinite limit at $ +\infty$ (see Figure fig example sandwich 1).

Corollary 4.6.3   Suppose that $ f$ is bounded in a (pointed) neighborhood of $ x_0$ and that $ \underset{x \rightarrow x_0}{\lim} g(x)=0$ . Then $ \underset{x \rightarrow x_0}{\lim} f(x)g(x)=0$ .

Example 4.6.4   We know that the sine function is bounded on $ \mathbb{R}$ , thus the function $ x \mapsto \sin \frac 1x$ is bounded on any pointed neighborhood of 0. Moreover $ \underset{x \rightarrow 0}{\lim}
x=0$ . Thus $ \underset{x \rightarrow 0}{\lim} x \cdot \sin \frac
1x =0$ . See Figure fig example sandwich 2.

Noah Dana-Picard 2007-12-28