Here is Prob. 15, Sec. 4.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $A \subseteq \mathbb{R}$, let $f \colon A \to \mathbb{R}$, and let $c \in \mathbb{R}$ be a cluster point of $A$. In addition, suppose that $f(x) \geq 0$ for all $x \in A$, and let $\sqrt{f}$ be the function defined for $x \in A$ by $\left(\sqrt{f}\right)(x) = \sqrt{f(x)}$. If $\lim_{x \to c} f$ exists, prove that $\lim_{x \to c} \sqrt{f} = \sqrt{ \lim_{x \to c} f}$.
My Attempt:
Let us put $$ L \colon= \lim_{x \to c} f(x). $$ Let us take any real number $\varepsilon > 0$.
First suppose that $L = 0$. Then there exists a real number $\delta > 0$ such that $$ \lvert f(x) - 0 \rvert < \varepsilon^2 $$ for all $x \in A$ which satisfy $$ 0 < \lvert x - c \rvert < \delta. $$ But as $f(x) \geq 0$ for all $x \in A$, so for all $x \in A$ for which $$ 0 < \lvert x-c \rvert < \delta, $$ we have $$ f(x) = \lvert f(x) \rvert = \lvert f(x) - 0 \rvert < \varepsilon^2, $$ and so $$\left\lvert \left(\sqrt{f}\right)(x) - 0 \right\rvert = \sqrt{ f(x) } < \varepsilon. $$ Thus it follows that $$ \lim_{x \to c} \left(\sqrt{f}\right)(x) = 0 $$ also.
Am I right?
Now suppose that $L \neq 0$. If $L$ were negative, then we would have $-L > 0$ and so we could find a real number $\delta > 0$ such that $$ \lvert f(x) - L \rvert < -L $$ for all $x \in A$ which satisfy $$ 0 < \lvert x-c\rvert < \delta. $$ Thus, for all $x \in A$ which satisfy $$ 0 < \lvert x-c\rvert < \delta, $$ we would obtain $$ L - (-L) < f(x) < L + (- L), $$ or $$ 2L < f(x) < 0. $$ But $f(x) \geq 0$ for all $x \in A$. Thus if $L$ were negative, then we would arrive at a contradiction. So $L$ must be (strictly) positive.
Am I right?
Now as $L > 0$ and as $f(x) \geq 0$ for all $x \in A$, so for every point $x \in A$, we find that $$ \begin{align} \left\lvert \left(\sqrt{f}\right)(x) - \sqrt{L} \right\rvert &= \left\lvert \sqrt{f(x)} - \sqrt{L} \right\rvert \\ &= \left\lvert \frac{ \left( \sqrt{f(x)} - \sqrt{L} \right) \left( \sqrt{f(x)} + \sqrt{L} \right) }{ \sqrt{f(x)} + \sqrt{L} } \right\rvert \\ &= \left\lvert \frac{ f(x) - L }{ \sqrt{f(x)} + \sqrt{L} } \right\rvert \\ &= \frac{ \left\lvert f(x) - L \right\rvert }{ \left\lvert \sqrt{f(x)} + \sqrt{L} \right\rvert } \\ &= \frac{ \left\lvert f(x) - L \right\rvert }{ \sqrt{f(x)} + \sqrt{L} } \\ &\leq \frac{ \left\lvert f(x) - L \right\rvert }{ \sqrt{L} } \\ &= \frac{ 1 }{ \sqrt{L} } \left\lvert f(x) - L \right\rvert. \tag{1} \end{align} $$ Now as $$ \lim_{x \to c} f(x) = L, $$ so we can find a real number $\delta_0 > 0$ such that $$ \lvert f(x) - L \rvert < \varepsilon \sqrt{L} $$ for all points $x \in A$ for which $$ 0 < \lvert x-c \rvert < \delta_0. $$ Thus from (1) we can conclude that, for all points $x \in A$ for which $$ \lvert x-c\rvert < \delta_0, $$ we end up with $$ \left\lvert \left(\sqrt{f}\right)(x) - \sqrt{L} \right\rvert \leq \frac{ 1 }{ \sqrt{L} } \left\lvert f(x) - L \right\rvert < \frac{ 1 }{ \sqrt{L} } \varepsilon \sqrt{L} = \varepsilon, $$ and hence we obtain $$ \left\lvert \left(\sqrt{f}\right)(x) - \sqrt{L} \right\rvert < \varepsilon. $$ Thus it follows that $$ \lim_{x \to c} \left(\sqrt{f}\right)(x) = \sqrt{L}. $$
Am I right?
Is this proof good enough? What is the presentation like?