# Cubic root of sequence translates into cubic root of limit?

I have this sequence:

$$(a_n) , n\in \mathbb N$$

$$\lim_{n\to \infty} a_n = a$$

Then I have this other sequence:

$$b_n = \sqrt{a_n}$$

$$\lim_{n\to \infty} b_n = \sqrt{a} ??$$

I need to know the reason behind the answer. Can you point me to the theorems or properties that solve this?

I'm thinking that maybe the solution is within the definition of limit, but I can't find it: $$\lim_{n\to \infty} a_n = a \iff \forall \varepsilon>0 , \exists n_0 : \forall n \ge n_0 , |a_n - a|< \varepsilon$$

• Why, yes, since the cubic root function is continuous. – Bernard Apr 12 '16 at 18:38

## 2 Answers

Yes. In this case, $\lim_{n\to\infty}b_{n}=\sqrt{a}$. This is because continuous functions preserve limits. I.e., if $f$ is continuous and $a_{n}\to a$, then $f(a_{n})\to f(a)$. The result follows in your case since $\sqrt{}$ is continuous.

• I understood your answer, thank you. – Joaquin Benitez Apr 12 '16 at 19:02

You can write

$$|\sqrt{a_{n}}-\sqrt{a}|=\frac{|a_{n}-a|}{\sqrt{a^2}+\sqrt{a_{n}^2}+\sqrt{aa_{n}}} \leq \frac{|a_{n}-a|}{\sqrt{a^2}} \quad (1)$$

Chose any $$c>0$$ $$\exists N(c \frac{\sqrt{a^2}}{2})$$ such that $n>N(c \frac{\sqrt{a^2}}{2})$

$$n>N(c\sqrt{a^2}) \Rightarrow |\sqrt{a_{n}}-\sqrt{a}| \leq c\frac{\sqrt{a^2}}{2}$$

$$(1) \quad \Rightarrow |\sqrt{a_{n}}-\sqrt{a}| \leq \frac{c}{2} <c$$