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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.

Andrew Falanga
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2 Answers2

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Hint:

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

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

RRL
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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? – Andrew Falanga Feb 23 '17 at 22:34
  • 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 $a_n \to L$. – RRL Feb 23 '17 at 23:07
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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.

Squirtle
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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. – Andrew Falanga Feb 22 '17 at 17:31
  • 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.

     – Squirtle Mar 01 '17 at 17:56