Proving the following limit statements I need to prove those


*

*If $f(x)\ge 0$ and $\lim_{x\to x_0}f(x)=L$, then $\lim_{x\to x_0}\sqrt{f(x)}=\sqrt{L}$.

*If $\lim_{x\to x_0}f(x)=L$, then $\lim_{x\to x_0}|f(x)|=|L|$.

*If $f(x)\ge g(x)$ for every $x$ in the neighborhood of $x_0$ (maybe not in $x_0$ itself) and if $\lim_{x\to x_0}f(x)=L_1, \,\lim_{x\to x_0}g(x)=L_2$, then $L_1\ge L_2$.


but I am really confused because I think I have a counter example for the first two :


*

*For the first:
$$f(x) =\begin{cases}0, & \text{if } x < 0 \\ 1, & \text{if } 0 \le x < 4\\ x^3, & \text{if } 4 \le x\end{cases}$$
In that way, if $x \to 4$, then the limit of $f(x)$ is $64$, but the limit of $\sqrt{4} = 2$ is $1$, and not $\sqrt{64} = 8$.

*Similar thing to the second one.
$$f(x) =\begin{cases}1, & \text{if } x \le 0 \\ x^2, & \text{if }0<x \end{cases}$$ In that way if $x \to -2$, then the limit of $f(x) = 1$, but the limit of $|f(x)| = 4$ and not $|1| = 1$. I am probably missing something here, yet I don't know what is it.

*As for the third one, I can see how it is true yet I am not sure how I can prove it.

 A: All three statements are correct; the following arguments comprise a "detailed" hint:
i)
If $L = 0$, then (formally) we have $|\sqrt{f(x)} - L| = \sqrt{f(x)} < \varepsilon$ iff $f(x) < \varepsilon^{2}$; suppose $L \neq 0$. Then for all suitable $x$ we have
$$
|\sqrt{f(x)} - \sqrt{L}| = \frac{|f(x) - L|}{\sqrt{f(x)} + \sqrt{L}},
$$
implying by assumption that there is some $\delta_{1} > 0$ such that $0 < |x-x_{0}| < \delta_{1}$ implies
$|f(x) - L| < L/2$, implying
$|f(x)| > L/2$, implying $\sqrt{f(x)} > \sqrt{L/2}$, implying
$$
\frac{|f(x) - L|}{\sqrt{f(x)} + \sqrt{L}} < \frac{|f(x) - L|}{\sqrt{L/2} + \sqrt{L}};
$$
for every $\varepsilon > 0$, by assumption there is some $\delta_{2} > 0$ such that $0 < |x-x_{0}| < \delta_{2}$ implies 
$$
|f(x) - L| < \varepsilon \bigg( \sqrt{\frac{L}{2}} + \sqrt{L}\bigg).
$$
ii)
Note that (formally) we have $||f(x)| - |L|| \leq |f(x) - L|$.
iii)
If $L_{1} < L_{2}$, then there is some $\delta_{1} > 0$ such that $0 < |x-x_{0}| < \delta_{1}$ implies 
$$
|g(x) - L_{2}| < \frac{L_{2}-L_{1}}{3}
$$ 
and there is some $\delta_{2} > 0$ such that $0 < |x-x_{0}| < \delta_{2}$ implies 
$$
|f(x) - L_{1}| < \frac{L_{2}-L_{1}}{3};
$$
hence $0 < |x - x_{0}| < \min \{\delta_{1}, \delta_{2}\}$ implies $g(x) > f(x)$, a contradiction.
A: For the first part of the question, observe that for any given $\epsilon > 0$, there exists $\delta_0 > 0$ such that $x \in (x_0-\delta_0, x_0+\delta_0) \Rightarrow |f(x)-L|< \epsilon \sqrt{L}$Thus for any given $\epsilon > 0$, if you select $\delta = \delta_0$, then $x \in (x_0 - \delta, x_0 + \delta)$ implies that $$ |\sqrt{f(x)}-\sqrt{L} | \sqrt{L} \le |\sqrt{f(x)}-\sqrt{L}| \left ( \sqrt{f(x)}+\sqrt{L} \right ) =|f(x)-L|< \epsilon \sqrt{L}$$,i.e.$|\sqrt{f(x)}-\sqrt{L}|<\epsilon$ for all $x \in (x_0-\delta,x_0+\delta)$
Similarly for the second one, see that $$||f(x)|-L|<|f(x)-L|$$ for the third one observe that since $f(x)-g(x)>0, \lim_{x \to x_0}f(x)-g(x) \ge 0$ and now since both the functions have limits at $x_0$, it follows that $L_1 \ge L_2$
