Okay, so heres the problem:

Suppose $\displaystyle f$ is the function satisfying $\displaystyle |f(x)| \leq |x|$ for all $\displaystyle x$. Show that $\displaystyle f$ is continuous at $\displaystyle 0$.

Okay, this is my work so far:

We need to show $\displaystyle \lim_{x \to 0} f(x) = f(0)$, I started by figuring the value of $\displaystyle f(0)$. To find that value, I proceeded with:

Since:

$\displaystyle x = 0$

and:

$\displaystyle |f(x)| \leq |x|$

then:

$\displaystyle |f(0)| \leq |0| \;\; \Rightarrow \;\; |f(0)| \leq 0$

but:

$\displaystyle |a| \geq 0 \;\;\;\; for \;\; all \;\; a$

So we must have:

$\displaystyle |f(0)| \geq 0$

But we also have the restriction that:

$\displaystyle |f(0)| \leq 0$

Therefore it must be true that:

$\displaystyle |f(0)| = 0 \;\; \Rightarrow \;\; f(0) = 0$

So, now i just need to show that:

$\displaystyle 0 < |x-0| < \delta \;\; \Rightarrow \;\; |f(x) - f(0)| < \epsilon$

but since $\displaystyle f(0) = 0$; we need only show:

$\displaystyle 0 < |x| < \delta \;\; \Rightarrow \;\; |f(x)| < \epsilon$

This is where i'm stuck. I betting the answer is right under my nose, and is really obivous, and I'm just not seeing it. I appreciate any help that anybody can give.