Find a way to define f(0) so that the function x sin (1/x) is continuous at 0

by RoRi

August 31, 2015

For $x \neq 0$ define

$f(x) = x \sin \frac{1}{x}.$

Give a value for $f(0)$ to make $f$ continuous at $x = 0$ .

We claim that if we define $f(0) = 0$ , then the function $f$ with this additional point defined is continuous at $x = 0$ .

Proof. Since $-1 \leq \sin \frac{1}{x} \leq 1$ for all $x \neq 0$ we know

$-x \leq x \sin \left( \frac{1}{x} \right) \leq x \qquad x \neq 0.$

Then since

$\lim_{x \to 0} -x = \lim_{x \to 0} x = 0$

we apply the squeeze theorem (Theorem 3.3 in Apostol) to conclude

$\lim_{x \to 0} x \sin \frac{1}{x} = 0.$

Therefore, by defining $f(0) = 0$ , we have extended $f$ to a function continuous at $0. \qquad \blacksquare$

Stumbling Robot