Prove that if $f$ is continuous and $f(x)=0$ for all $x<0$ then $f(0)=0$ 
Prove that if $f$ is continuous in $\mathbb{R}$ and $f(x)=0$ for all $x<0$ then $f(0)=0$.

My attempt: $f(x)=0$ for all $x<0$ can be written as for all $\varepsilon>0$ it is $f(x-\varepsilon)=0$.
Letting $x\to0^-$ both sides of the equality leads to
$$\lim_{x\to0^-} f(x-\varepsilon)=0$$
Since by hypothesis $f$ is continuous the limit on the left hand side can be evaluated by simply substituting $x=0^-$ in $f$, so we have
$$f(-\varepsilon)=0$$
Since it is true for all $\varepsilon>0$ it is true if we let $\varepsilon \to 0^+$: letting $\varepsilon \to 0^+$ we obtain, using again the continuity of $f$, the result $f(0)=0$.
Is this correct? Thanks.
 A: Sure, it's correct, but you used continuity of $f$ twice, when you only had to use it once. We have:
\begin{align}
f(0) = \lim_{x \to 0}f(x) = \lim_{x \to 0^-}f(x) = \lim_{x \to 0^-}(0) = 0
\end{align}
First is by continuity, second is because the limit exists, third is by hypothesis, and the last is obvious.
A: You are doing the work twice, i.e., you consider $x\to 0^-$ and in a second step $\epsilon\to 0^+$. One of these suffices.
So what you basically  use is that if $x_n$ is a convergent sequence in the domain of $f$, then by continuity
$$f(\lim_{n\to\infty}x_n)=\lim_{n\to\infty}f(x_n). $$
Why not use this directly with $x_n=-\frac1n$?
$$f(0)=f(\lim_{n\to\infty}-\tfrac1n)=\lim_{n\to\infty}f(-\tfrac 1n)=\lim_{n\to\infty}0=0. $$
A: Rephrasing:
Since f is continuos at 0:
$\lim_{x \rightarrow 0^{-}}f(x)=\lim_{x \rightarrow 0^{+}}=$
$\lim_{x \rightarrow 0}f(x).$
For any sequence $x_n <0$, with 
$\lim_{n \rightarrow \infty}x_n=0$ we have $f(x_n)=0.$
$\lim_{n \rightarrow \infty}f(x_n)=0$, and we are done.
