# Show that this piecewise function is continuous at 0

## Homework Statement

##f(x) = x \sin (\frac{1}{x})## for ##x \ne 0## and ##f(0) = 0##. Prove that this function is continuous at 0.

## The Attempt at a Solution

First, I need to look at the quantity ##|f(x) - f(0)|##. However, I am not completely sure how to proceed. I would think that we substitute ##x \sin (\frac{1}{x})## for ##f(x)##, but is this justified? Why can we guaruntee in this case that ##x## won't assume the value of 0 and hence be 0 rather than ##x \sin (\frac{1}{x})##?

Mark44
Mentor

## Homework Statement

##f(x) = x \sin (\frac{1}{x})## for ##x \ne 0## and ##f(0) = 0##. Prove that this function is continuous at 0.

## The Attempt at a Solution

First, I need to look at the quantity ##|f(x) - f(0)|##.
Depending on how rigorous your argument needs to be, you might not need to look at this difference.
One definition for the continuity of function a at a number ##x_0## in its domain is this: ##\lim_{x \to x_0}f(x) = f(x_0)##

Mr Davis 97 said:
However, I am not completely sure how to proceed. I would think that we substitute ##x \sin (\frac{1}{x})## for ##f(x)##, but is this justified? Why can we guaruntee in this case that ##x## won't assume the value of 0 and hence be 0 rather than ##x \sin (\frac{1}{x})##?
Because in the definition you're dealing with a "punctured neighborhood" around 0.

I.e., if ##0 < |x - 0| < \delta##

member 587159

## Homework Statement

##f(x) = x \sin (\frac{1}{x})## for ##x \ne 0## and ##f(0) = 0##. Prove that this function is continuous at 0.

## The Attempt at a Solution

First, I need to look at the quantity ##|f(x) - f(0)|##. However, I am not completely sure how to proceed. I would think that we substitute ##x \sin (\frac{1}{x})## for ##f(x)##, but is this justified? Why can we guaruntee in this case that ##x## won't assume the value of 0 and hence be 0 rather than ##x \sin (\frac{1}{x})##?

You can't substitute that value if you want to use epsilon delta definition.

You can however consider two cases: namely x = 0 and ##x \neq 0##. The former case gives always 0, so you can safely assume ##x \neq 0## and then you are allowed to plug in the expression.

It then boils down to estimating ##|x \sin(1/x) - 0|## for ##x \neq 0##, given that x is small, which ought to be straightforward.