Verify epsilon-delta continuity for $\sqrt x, (x\ge 0)$ at $x=4$ and $x=100$

Question

Define $f(x)=\sqrt{x}$ for all $x\geq 0$. Verify the $\epsilon,\delta$ criterion for continuity at x=4 and at x=100. Hint: first show that for $x\geq0$, $x_0\gt0$, $|\sqrt{x}-\sqrt{x_0}|\leq|x-x_0|/\sqrt{x_0}$.

For my proof (just for $x=4$), I used that $\forall x \geq0, \sqrt{x} \leq x$. So, I let $\delta = \epsilon$, and then I let $x \in \mathbb{R}$ be such that $|x-4| \lt \delta$. Then, $|\sqrt{x}-\sqrt{4}|\leq|x-4|\lt \delta = \epsilon$. Is this a valid statement? I feel a little "iffy" about the validity of going from my $|\sqrt{x}-\sqrt{4}|$ to $|x-4|$.

I am also confused about the hint that the problem gave. Why would I need to prove that first? (In class, we just do the problems applying the $x_0$ value immediately.)

Answer 1

Guide:

• As mentioned in the comment, $\forall x \ge 0, \sqrt{x} \le x$ is not true.

• $|\sqrt{\frac14} - \sqrt{\frac1{16}} | =\frac12-\frac14=\frac14=\frac{4}{16} $ but $\frac14 - \frac1{16}=\frac{3}{16}$

• If you have the hint, then your proof can be as follows.

For example, let $x_0=4$, then

If $0< |x-4|< \delta$, then $|\sqrt{x}-\sqrt{4}| \le \frac{|x-4|}{\sqrt4}\le \frac{\delta}2$ and you can choose your $\delta$ in terms of your $\epsilon$ easily.

• Useful tools to prove the hint: $a^2-b^2 = (a-b)(a+b)$ and also if $a>0, b \ge 0$, then $\frac{1}{a+b} \le \frac1a$ .