$f(x)$ has a limit, prove that $\sqrt{f(x)}$ has a limit

Question

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 !

Answer 1

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).

Answer 2

Hint: The function $x \mapsto\sqrt{x}$ is continuous for $x> 0$.

Answer 3

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 $x$.