$f(x)$ has a limit, prove that $\sqrt{f(x)}$ has a limit The question is :

Let $f$ be a positive function defined on an interval $[a,\infty)$, such that $\lim\limits_{x\to\infty} f(x)=0$. Prove that $\lim\limits_{x\to\infty} \sqrt{f(x)}=0$.

If $f(x)$ has a limit it means there is an $\varepsilon > 0$ and an $M>0$ s.t. for every $x>M$, $|f(x) - l| < \varepsilon.$
It seems kind of obvious that it is true yet I can't find a way to prove it.
Thanks in advance !
 A: Hint: The function $x \mapsto\sqrt{x}$ is continuous for $x> 0$.
A: The problem is in fact much more simpler than you probably think: If $\varepsilon > 0$, then
$\sqrt{f(x)} < \varepsilon$ iff $f(x) < \varepsilon^{2}$ for all suitable $x$.
A: These type of problems always boil down to the choice of $\varepsilon$.
$$\lim_{x\to\infty}f(x) = 0\Leftrightarrow \forall\varepsilon >0, \exists K>0:\forall x\left (x>K\Rightarrow |f(x)|<\varepsilon\right )$$
Assuming $f(x)\geq 0$ for every $x\in [a,\infty)$. Objective is to show that $x>K\Rightarrow \sqrt{f(x)}<\varepsilon $  (square root is non-negative, no need for absolute value signs).
Assuming $x>K$, then $f(x)<\varepsilon$. Is it then true, that also $\sqrt{f(x)}<\sqrt{\varepsilon}$? Yes, it is.
Therefore, let $\varepsilon := \varepsilon ^2$. Then
$$x>K\Rightarrow f(x)<\varepsilon ^2 \Rightarrow \sqrt{f(x)}<\sqrt{\varepsilon ^2} = |\varepsilon| =\varepsilon$$
Prove the same thing in the general case, now :)
You would have to convince the reader that $x<A$ implies $\sqrt{x} < \sqrt{A}$, assuming $x,A$ are both non-negative.
EDIT:
To understand limit proofs you need a strong base in discrete mathematics (for anything in math, really). You have to be able to read and understand implications. (by the sound of things you are doing introductory analysis).
