# Lim a_n = L => lim sqrt(a_n) = sqrt(L)

Let $$\{a_n\}$$ be a sequence of positive numbers. If $$\lim a_n = A$$, prove that $$\lim \sqrt{a_n} = \sqrt{A}$$.

I thoght that $$f(x) = \sqrt{x}$$ is a continuous function, and composing a sequence that has a limit with a continuous function must have a limit that is just evaluating the previous limit in the function.

But I think I'm not supposed to use that fact, as the concept of limit of sequence comes previously to the definition of continuous function in the course.

It seems trivial, but still can't find how to start. Any help?

What is your definition of a limit?

What is your definition of a limit?

Hi kru_,
Isn't this standard for sequences?

A sequence $$a_n$$ of real numbers has limit $$L \in \mathbb{R}$$ iff for every $$\epsilon > 0$$ there exists $$n_0 \in \mathbb{N}$$ such that for all $$n \geq n_0$$ one has $$|a_n - L| < \epsilon$$

So you know $|a_n - A| < \epsilon$.

I think the key hint here is that $|\sqrt{a_n} - \sqrt{A}| = \frac{|a_n - A|}{|\sqrt{a_n} + \sqrt{A}|}$

Thank you, but I still can't see how to end it. Most that can see is that

$| \sqrt{a_n} - \sqrt{A}| < \frac{\epsilon}{|\sqrt{a_n} + \sqrt{A}|}$

And I'm not sure if it's still valid if $\lim_{n \to \infty} a_n = 0$

Thank you, but I still can't see how to end it. Most that can see is that

$| \sqrt{a_n} - \sqrt{A}| < \frac{\epsilon}{|\sqrt{a_n} + \sqrt{A}|}$

And I'm not sure if it's still valid if $\lim_{n \to \infty} a_n = 0$

You are right that you will need to handle a separate case if the limit is 0. This should be a pretty straightforward case, however. What does the definition of the limit look like if A = 0?

Focusing on the general case, where A is not 0:

You know that $|a_n - A| \leq \epsilon$. The crucial component of this knowledge is that this remains true for all positive epsilon. It remains true for $2\epsilon, 3\epsilon, \frac{3\epsilon}{7}, 4\pi\epsilon, \epsilon^2,..., A\epsilon^2, \sqrt{A}\epsilon$, etc.

Does this make sense?

Another thing to consider is what you can do to the term $\frac{|a_n - A|}{|\sqrt{a_n} + \sqrt{A}|}$ to make it bigger.