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

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

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

One comment

  1. Dimitris says:

    Just a detail… Since x can be either positive or negative we get two inequalities
    -x <= ….. <= x and x<= …. <= -x and we have to apply the Squeeze Principle two times, one for each side limit.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):