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This is a homework problem for which I think I've missed the point or have incorrectly done the proof (or both). There are two parts to the problem: Let $(a_n)$ be a sequence with $a_n \ge 0,$ for all $n$.

Part 1:
Suppose that $a_n \rightarrow 0$. Show that $\sqrt{a_n}\rightarrow 0$. I recognized that these are actually two different sequences. Here is my proof.
Proof
Let $(b_n)=\sqrt{a_n}$ and $\forall a\in (a_n), a\ge 0$.
By assumption, $(a_n) \rightarrow 0$.
By definition of Square Root, $\forall b_n, b_n < a_n$.
So, since $b_n < a_n$ and $(a_n) \rightarrow 0, (b_n) \rightarrow 0 \square$.

Part 2:
Suppose that $a_n \rightarrow L$. Show that $\sqrt{a_n}\rightarrow \sqrt{L}$.

Assuming I did things correctly in part 1, shouldn't this be a virtual ditto? My teacher gave this hint which makes me think I'm way off base:
you can assume that $L\ne 0$. Use that $\sqrt{x} - \sqrt{y} = \frac{x-y}{\sqrt{x}+\sqrt{y}}$.

Since my proof in part 1 didn't use anything like this, I'm assuming I'm off in the weeds. Since this homework hasn't yet been graded, I'll need hints rather than solutions. Thanks.

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    In part 1, $\sqrt{a_n}>a_n$ for $0< a_n<1$. For example, $\sqrt{1/4}=1/2>1/4$. In part 2, I think you mean $\sqrt{a_n}\to\sqrt{L}$. – 2017-02-22
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    @ForgotALot you're correct. I'm fixing that now. – 2017-02-23

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Hint:

For an all-encompassing proof with $L \geqslant 0$ note that

$$|\sqrt{a_n} - \sqrt{L}|^2 \leqslant |\sqrt{a_n} - \sqrt{L}||\sqrt{a_n} + \sqrt{L}| = |a_n - L|$$

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    I understand what you're saying. What I don't understand is the "magic" or the why you know to do this. Is it simply going back to the definition of *convergence*? That is, namely, to show convergence, we must show that at some $N\in\mathbb{N}, \forall n\in \mathbb{N}$ such that $n\ge N, |a_n - L| \lt \epsilon$. Is this why? – 2017-02-23
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    Yes and more to the point we have to show $|\sqrt{a_n} - \sqrt{L}| < \epsilon$ for sufficiently large $n$. What we are given is that $a_n \to L$ so we know we can make $|a_n - L|$ small. In the end we have to relate $|\sqrt{a_n} - \sqrt{L}|$ to $|a_n - L|$. Using what I showed, you can see that if $|a_n - L| < \epsilon^2$ then $|\sqrt{a_n} - \sqrt{L}| < \epsilon$. The first condition can be met for sufficiently large $n$ since $a_n \to L$. – 2017-02-23
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For the first one it is NOT true that $\sqrt{x_n} < x_n$ but the limit is still zero all the same. Are you allowed to use the fact that you can "pass a limit through a continuous function"? If so... then $$\lim_{n\to \infty} g(x_n) = g(\lim_{n\to \infty} x_n) = g(0)$$ where $g(x)=\sqrt{x}$ and we know that $g(0)=0$. Please let me know if you are not aware of this result or cannot use it.

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    I had suspected that my reasoning was incorrect. I do not know if this would not be permitted. However, I didn't think of using such a construct. – 2017-02-22
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    http://math.stackexchange.com/questions/560307/prove-that-sqrtx-is-continuous-on-its-domain-0-infty This proves that $f(x)=\sqrt{x}$ is continuous on $[0,\infty)$ and your textbook should have a proof that the first equals sign is true in my centred equation. – 2017-03-01