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 $$|f(x)-L|
$|\sqrt{f}+\sqrt{L}|>\sqrt{L/2}+\sqrt{L}$. Denote $\sqrt{L/2}+\sqrt{L}$ by $M$. Hence $1/{|\sqrt{f}+\sqrt{L}|}<1/M$ for all $0<|x-x_0|<\delta_1$.
Now we can show that desired limit:
Let $\epsilon>0$, then there would be some $\delta_2>0$ such that if $0<|x-x_0|<\delta_2$, we must have$$|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 $\lim_{x\to x_0}\sqrt{f}=\sqrt{L}$.
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