Prove that $\lim_{x\to a}f(x)=L$ implies $\lim_{x\to a}\sqrt{f(x)}=\sqrt L$

Question

If $\displaystyle\lim_{x\to a}f(x)=L>0.$ Prove $\displaystyle\lim_{x\to a}\sqrt{f(x)}=\sqrt{L}$.

I know that we have:

$$\left|\sqrt{f(x)}-\sqrt{L}\right|=\left|\frac{f(x)-L}{\sqrt{f(x)}+\sqrt{L}}\right|<\left|\frac{f(x)-L}{L}\right|<|f(x)-L|<\varepsilon\text{.}$$

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Proof: Given $\varepsilon>0$. Let $\delta=\varepsilon$. Assume $\displaystyle\lim_{x\to a}f(x)=L>0$. So there exists a $\delta>0$ such that

$$\text{if }\; 0<|x-a|<\delta=\varepsilon\text{, then }\;|f(x)-L|<\delta=\varepsilon\text{.}$$

So, If $0<|x-a|<\delta$ then

$$\left|\sqrt{f(x)}-\sqrt{L}\right|=\frac{|f(x)-L|}{\sqrt{f(x)}+\sqrt{L}}\le \frac{|f(x)-L|}{L}<|f(x)-L|<\delta=\varepsilon\text{.}$$

Right or am I wrong?

Answer 1

It might help to use different variables.

Since $\lim_{x\to a} f(x) = L > 0$, we know that:

$\forall \varepsilon_1 > 0$, $\exists \delta_1 > 0$ such that if $0 < |x - a| < \delta_1$, then $|f(x) - L| < \varepsilon_1$

Now given any $\varepsilon > 0$, let $\delta > 0$ be the $\delta_1$ corresponding to taking $\varepsilon_1 = \varepsilon\sqrt{L}$. Then observe that if $0 < |x - a| < \delta = \delta_1$, then: \begin{align*} \left|\sqrt{f(x)} - \sqrt{L} \right| &= \left|\frac{f(x) - L}{\sqrt{f(x)} + \sqrt{L}}\right| \\ &= \frac{|f(x) - L|}{\sqrt{f(x)} + \sqrt{L}} \\ &\leq \frac{|f(x) - L|}{\sqrt{L}} \\ &< \frac{\varepsilon\sqrt{L}}{\sqrt{L}} \\ &= \varepsilon \end{align*} Hence, it follows that $\lim_{x\to a} \sqrt{f(x)} = \sqrt{L}$, as desired. $~~\blacksquare$