If ${S_{n}}$ is a sequence of positive number converging to 0,Show that $\sqrt{S_{n}}$ converging to 0

Question

Please check my proof Because $\sqrt{S_{n}}$ is composite function which outer function is$\sqrt{x}$ and $S_{n}$

We can prove by consider $\lim_{n \to\infty }\sqrt{\lim_{n \to \infty }S_{n}}$

First consider converge of $S_{n}$ to 0

Given $\epsilon > 0 we must find N such that n>N

$$|S_{n}-L|<\epsilon $$

in this case L is 0 then

$$S_{n}< \epsilon $$

$$\frac{1}{\epsilon }<\frac{1}{S_{n}} $$

Choose $N\geq \frac{1}{\epsilon }$

then $\frac{1}{\epsilon }\leq N< n$

then limit is 0

Now consider $\lim_{n \to \infty }\sqrt{0}$

$$|0-0|<\epsilon $$

$$|0|<\epsilon $$

Therefore $\sqrt{S_{n}} $ converge to 0

Answer 1

You have confused the question. You are given that $S_n \to 0$, and you have to prove that $\sqrt {S_n} \to 0$. The proof you have given is not entirely clear, either.

To do this, let $\epsilon > 0$. We know there is $N \in \mathbb N$ such that $n > N \implies S_n < \epsilon^2$.But then $S_n < \epsilon^2 \implies \sqrt {S_n} < \epsilon$ for all $n > N$. Hence, it follows that $S_n \to 0$.

Note that this actually shows that the (positive) square root is continuous at the point $0$.