How do I show that if $(a_n)$ is a sequence of nonnegative numbers converging to $L$, then $L \ge 0$? 
Let $(a_n)$ be a sequence such that $a_n \geq 0 $ for all $n$ and
  $\lim a_n = L $. Prove very carefully that $L \geq 0$

This exercise I am trying to solve. Here is my ${\bf attempt:}$
Since $(a_n)$ converges to $L$, then with $\epsilon = |L| > 0$ we can obtain an $N$ so that for all $n > N$ one has 
$$  a_n - L < |L| $$
and thus $0 \leq a_n < L + |L| $. If $L<0$, then $L+|L|=0$ and we obtain $0<0$ which is certainly not true! therefore $L \geq 0$ and we are done.
Is this a correct proof? Any criticism would be greatly appreciated
 A: Yes, it is correct. Here is how I would have done it. Suppose $L<0$. Since $a_n\to L$, there exists $N$ such that for all $n>N$ we have $|a_n-L|<|L|/2$. Then
$$
a_n<L+|L|/2<0,
$$
contradicting $a_n\geq0$. Thus $L\geq0$. 
A: Here I provide an alternative way to solve it just for the sake of curiosity.
Lemma
If $a_{n}\geq b_{n}$ and $(a_{n})_{n=m}^{\infty}\to a$ and $(b_{n})_{n=m}^{\infty}\to b$, then $a\geq b$.
Proof
According to the definition of limits for every $\varepsilon > 0$, there are $N_{1}\geq m$ and $N_{2}\geq m$ such that
\begin{align*}
\begin{cases}
n\geq N_{1}\\\\
n\geq N_{2}
\end{cases} \Longrightarrow
\begin{cases}
|a_{n} - a| < \varepsilon\\\\
|b_{n} - b| < \varepsilon
\end{cases} & \Longrightarrow b - \varepsilon < b_{n} \leq a_{n} < a +\varepsilon\\\\
& \Longrightarrow a - b + 2\varepsilon > 0
\end{align*}
If we assume that $a < b$ and choose $\displaystyle\varepsilon = \frac{b-a}{2}$, then we can take $N = \max\{N_{1},N_{2}\}$ such that
\begin{align*}
a - b + b - a = 0 > 0
\end{align*}
which is impossible. Therefore $a \geq b$.
Solution
At your case, $(a_{n})_{n=m}^{\infty}$ converges to $L$, $b_{n} = 0$ is the null sequence and $a_{n} \geq b_{n}$.
Hopefully it is useful.
A: You should add "We assume $L < 0$ and aim for a contradiction". Then your proof is complete.
Here is an alternative proof:
If $L < 0$, there is $m$ such that $a_m < 0$ (choose $\epsilon = -L/2$ in the definition of limit), impossible by our assumption.
