real analysis - Prove: if $limlimits_{x rightarrow x_0} f(x) = L >0$, then $limlimits_{x rightarrow x_0} sqrt{f(x)}= sqrt{L}$

Prove: if $\lim\limits_{x \rightarrow x_0} f(x) = L >0$, then $\lim\limits_{x \rightarrow x_0} \sqrt{f(x)}= \sqrt{L}$ (using the original definition of limit)

We assume that $f$ is defined in some deleted neighborhood of $x_0$, for every $\epsilon>0$ , there is a $\delta>0$ such that:

$$|f(x)-L|<\epsilon$$
and
$$ 0<|x-x_0|<\delta$$

As $L>0$,

$$|f(x)-L|
=| \sqrt{f} \sqrt{f} - \sqrt{L} \sqrt{L}|
= |\sqrt{f} - \sqrt{L}| \cdot |\sqrt{f} + \sqrt{L}|

<\epsilon$$

It follows that for the same $\delta>0$, the following statement is true:
$$|f(x)-L|<|\sqrt{f} - \sqrt{L}|< \frac{\epsilon}{ |\sqrt{f} + \sqrt{L}|} < \epsilon$$

>

It shows that if $\sqrt{f}$ is defined on some deleted neighborhood of $x_0$,there is a

$\epsilon>0$ , there is a $\delta>0$ such that:

$$|\sqrt{f}-\sqrt{L}|<\epsilon$$
and
$$ 0<|x-x_0|<\delta$$

Am I having the right reasoning? Can it be improved?

Any input is much appreciated.

Answer

${\epsilon}/{|\sqrt{f}+\sqrt{L}|}$ does not necessarily have to be smaller than $\epsilon$ because $|\sqrt{f}+\sqrt{L}|$ could be lesser than $1$. Instead, look for a bound for $1/{|\sqrt{f}+\sqrt{L}|}$ :

Since $L>0,$ so $L/2>0$, and there would be (by definition of the limit of $f$ at $x_0$) some $\delta_1>0$ such that if $0<|x-x_0|<\delta_1$, we must have $$|f(x)-L|$$\Rightarrow -L/2So we have $f(x)>L/2>0$ for all $x$ satisfying $0<|x-x_0|<\delta_1$. Then
$|\sqrt{f}+\sqrt{L}|>\sqrt{L/2}+\sqrt{L}$. Denote $\sqrt{L/2}+\sqrt{L}$ by $M$. Hence $1/{|\sqrt{f}+\sqrt{L}|}<1/M$ for all $0<|x-x_0|<\delta_1$.

Now we can show that desired limit:

Let $\epsilon>0$, then there would be some $\delta_2>0$ such that if $0<|x-x_0|<\delta_2$, we must have$$|f(x)-L|<\epsilon$$
Take $\delta=\min(\delta_1,\delta_2)$, then if $x$ is any number satisfying $0<|x-x_0|<\delta$, we must have$$|\sqrt{f}-\sqrt{L}|<\epsilon/|\sqrt{f}+\sqrt{L}|<\epsilon/M$$

Since $M$ is fixed and $\epsilon$ is arbitrary, so $\lim_{x\to x_0}\sqrt{f}=\sqrt{L}$.