Let $\{a_n\}$ be a sequence such that $\forall n \in \mathbb{N}: a_n \geq 0$, and $a_n$ converges to $a \geq 0$.
Prove that $$\lim_{n \to \infty} \sqrt{a_n} = \sqrt{a}$$ I tried using the definition of limit and the sandwich theorem but it doesn't work.
It suffices to see that $$\left|\sqrt {a_n} -\sqrt a\right|=\frac {\left|a_n-a\right|} {\sqrt {a_n} +\sqrt a}\leq \frac {\left|a_n-a\right|} {\sqrt a}<\varepsilon$$ if $|a_n-a|< \varepsilon \sqrt a$.
There could also be a case in which $a = 0$ and then we cannot assume $|a_n-a|< \varepsilon \sqrt a$.
When $a = 0$ there exists $N$ such that for all $n > N$ $|a_n| < \varepsilon ^2$ and then $$|\sqrt{a_n}| = \sqrt{a_n} < \varepsilon \iff a_n < \varepsilon ^2 $$ and we showed the rhs is true $\forall n > N$
Since $a_n\to a$ exists $n_0$ such that for any $n\ge n_0$ we have $\left|a_n-a\right|<\epsilon\sqrt a$ and therefore
$$|\sqrt {a_n} - \sqrt a|=\left|\frac{a_n-a}{\sqrt {a_n} + \sqrt a}\right|=\frac{\left|a_n-a\right|}{\sqrt {a_n} + \sqrt a}\le\frac{\left|a_n-a\right|}{ \sqrt a}<\epsilon$$
Hint: There’s a more general statement. If $f$ is continuous, and $\lim_{n\to\infty}a_n=L$, then $$\lim_{n\to\infty}f\left(a_n\right)=f(L).$$ You can prove this by combining the definition of the limit with the definition of continuity.
