Show that this piecewise function is continuous at 0

[Mr Davis 97]

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.

Homework Equations

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]

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.

Homework Equations

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: $\lim_{x \to x_0}f(x) = f(x_0)$

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.

Homework Equations

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.