Simple epsilon-delta proof. The square root in 4

Question

Is this correct?

Show that $f(x)=\sqrt{x}$ is continous at the point $4$.

Choose $\epsilon = \frac{\delta}{2}$:

Then $|\sqrt{x}-2| < \frac{\sqrt{x}+2}{2}|\sqrt{x}-2| = \frac{|x-4|}{2} < \frac{\delta}{2} = \epsilon$.

Answer 1

You don't choose $\epsilon.$ Given $\epsilon>0$, you need to choose $\delta$ such that $$|x-4|<\delta\Rightarrow |f(x)-f(4)|=|\sqrt{x}-2|<\epsilon.$$

The following is a possible way to write the proof:

Let $\epsilon>0$ be given. Choose $\delta=\min(4,2\epsilon).$ Then if $|x-4|<\delta,$ one has $$|\sqrt{x}-2|<\frac {\sqrt{x}+2}2 |\sqrt{x}-2|=\frac {|x-4|}2<\frac{\delta}2\leq \epsilon,$$ as required.

Here it is necessary to make $\delta$ small enough so that $x\leq 0$ won't occur. For example, if $x<0$, $\sqrt{x}$ is undefined over the reals, and when $x=0$, the first inequality right above is not true.