Prove lim as x goes to a of sqrt( f(x)) equals sqrt(L)

If f(x) >= 0 for all x in a deleted neighborhood of a and lim as x goes to a of f(x) equals L.

Let ε > 0 be given.

Since lim(x→a) f(x) = L, there exists δ > 0 such that
0 < |x - a| < δ ==> |f(x) - L| < ε√L (and f(x) ≥ 0).

Therefore, 0 < |x - a| < δ implies that
|√f(x) - √L|
= |√f(x) - √L| |√f(x) + √L| / |√f(x) + √L|, using conjugates
= |f(x) - L| / (√f(x) + √L)
≤ |f(x) - L| / (0 + √L)
< (ε√L) / √L
= ε.

In other words, lim(x→a) √f(x) = √L, as required.

I hope this helps!

Assume L>0, and that f(x)>=0 in a deleted nieghborhood of zero,
since otherwise we are dealings in complex numbers.
Then
Sqrt[f(x)]-sqrt(L)
= { Sqrt[f(x)]-sqrt(L) }{Sqrt[f(x)]+sqrt(L)}/{ Sqrt[f(x)]+sqrt(L)}
= { f(x)-L }/{ Sqrt[f(x)]+sqrt(L)}

Now Sqrt[f(x)]+sqrt(L)>=sqrt(L)>0. Hence

|Sqrt[f(x)]-sqrt(L)|<|f(x)-L|/Sqrt(L).

Now given e>0 there exists delta>0 such that
|x-a| |f(x)-L|
|x-a||Sqrt[f(x)]-sqrt(L)|<|f(x)-L| /sqrt(L)
which proves limit as x->a of sqrt[f(x)]=sqrt(L).

Hope this helps