Let $\lim_{x \rightarrow a} f(x) = A$ . Prove $\lim_{x \rightarrow a} \sqrt{f(x)} = \sqrt{A}$. Help?

Question

I've been asked to prove this using the arguments $\space \varepsilon , \delta$ .

Now I know that for $\space \varepsilon >0 \space$ there exists a $\space \delta>0 \space$ such that:

$|x-a|<\delta \space \Rightarrow \space |f(x)-A|<\varepsilon$

As intuitively obvious as it is, I just can't seem to get to this:

$|\sqrt{f(x)} - \sqrt{A}|<\varepsilon$

I mean, I know this is ture intuitively, but can't prove it.

Thank you.

Answer 1

We assume that $A> 0$. For any given $\epsilon>0$, there exists a $\delta>0$ such that

$$|f(x)-A|<\sqrt{A}\epsilon$$

whenever $0<|x-a|<\delta$. Therefore, for the same $\epsilon>0$

$$\begin{align} \left|\sqrt{f(x)}-\sqrt{A} \right|&=\left|\frac{f(x)-A}{\sqrt{f(x)}+\sqrt{A}} \right|\\\\ &\le \frac{|f(x)-A|}{\sqrt{A}}\\\\ &<\epsilon \end{align}$$

whenever $0<|x-a|<\delta$. And we are done!