The Pinching or Sandwich Theorem

As a motivation let us consider the function

$$f(x) = x^2 \sin\left(\frac{1}{x}\right)\;\cdot$$

When x get closer to 0, the function $$g(x)=\sin\left(\frac{1}{x}\right)$$ fails to have a limit. So we are not able to use the basic properties discussed in the previous pages. But we know that this function $$g(x)=\sin\left(\frac{1}{x}\right)$$ is bounded below by -1 and above by 1, i.e.

$$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$

for any real number x. Since $x^2 \geq 0$, we get

$$- x^2 \leq x^2\;\sin\left(\frac{1}{x}\right) \leq x^2 \;.$$

Hence when x get closer to 0, x² and -x² become very small in magnitude. Therefore any number in between will also be very small in magnitude. In other words, we have

$$\lim_{x \rightarrow 0}\quad x^2\;\sin\left(\frac{1}{x}\right) = 0\;.$$

This is an example for the following general result:

Theorem: The "Pinching" or "Sandwich" Theorem Assume that

$$h(x) \leq f(x) \leq g(x)$$

for any x in an interval around the point a. If

$$\lim_{x \rightarrow a} h(x) = L\;\; \mbox{and}\;\; \lim_{x \rightarrow a} g(x)= L\;,$$

then

$$\lim_{x \rightarrow a} f(x) = L\;.$$

Example. Let f(x) be a function such that $\vert f(x)\vert \leq M$, for any $x \neq 0$. The Sandwich Theorem implies

$$\lim_{x \rightarrow 0} x\;f(x) = 0\;.$$

Indeed, we have

$$\vert x\; f(x)\vert \leq M \vert x\vert$$

which implies

$$-M \vert x\vert \leq x\; f(x) \leq M \vert x\vert$$

for any $x \neq 0$. Since

$$\lim_{x \rightarrow 0} M\;\vert x\vert = 0\;\;\mbox{and} \;\; \lim_{x \rightarrow 0} -M\;\vert x\vert = 0\;,$$

then the Sandwich Theorem implies

$$\lim_{x \rightarrow 0} x\;f(x) = 0\;.$$

Exercise 1. Use the Sandwich Theorem to prove that

$$\lim_{x \rightarrow a} \sqrt{x} = \sqrt{a}$$

for any a > 0.

Answer.

Exercise 2. Use the Sandwich Theorem to prove that

$$\lim_{x \rightarrow 0} x \cos\left(\frac{1}{x}\right) = 0\;.$$

Answer.

Exercise 3. Consider the function

$$f(x) = \left\{ \begin{array}{lll} 1 + 2x^2 &\mbox{if $x$ is r... ...\\ 1 + x^4 &\mbox{if $x$ is irrational.}\\ \end{array}\right.$$

Use the Sandwich Theorem to prove that

$$\lim_{x \rightarrow 0} f(x) = 1\;.$$

Answer.

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