I was asked to prove, using the definition of limit, that
$$\lim\limits_{x \to 4}{\frac{\sqrt{x}-2}{x-4}} = \frac{1}{4}$$
I tried doing it but I am not completely sure my prove is correct. I am publishing it here so that more knowledgeable people can check it for me, maybe pointing any pitfalls.
I need to prove that for any $\epsilon > 0$ there is a $\delta > 0$ such that if $0 < \left|x - 4\right| < \delta$ then $\left|\frac{\sqrt{x}-2}{x-4} - \frac{1}{4}\right| < \epsilon$
First part: Tying to find a suitable $\delta$, given an $\epsilon$
$$\left|\frac{\sqrt{x}-2}{x-4} - \frac{1}{4}\right| < \epsilon$$ $$\left|\frac{4(\sqrt{x}-2) - (x-4)}{4(x-4)}\right| < \epsilon$$ $$\left|\frac{4\sqrt{x} - 8 - x + 4}{4(x-4)}\right| < \epsilon$$ $$\left|\frac{4\sqrt{x} - x - 4}{4(x-4)}\right| < \epsilon$$ $$\left|\frac{-x + 4\sqrt{x} - 4}{4(x-4)}\right| < \epsilon$$ $$\left|\frac{-(x - 2\cdot\sqrt{x}\cdot 2 + 4)}{4(x-4)}\right| < \epsilon$$ $$\left|\frac{-(\sqrt{x} - 2)^2}{4(\sqrt{x}+2)(\sqrt{x}-2)}\right| < \epsilon$$ $$\left|\frac{-(\sqrt{x} - 2)}{4(\sqrt{x}+2)}\right| < \epsilon$$ $$\left|-\frac{1}{4}\right| \left|\frac{\sqrt{x} - 2}{\sqrt{x}+2}\right| < \epsilon$$ $$\frac{1}{4} \left|\frac{\sqrt{x} - 2}{\sqrt{x}+2}\right| < \epsilon$$ $$\left|\frac{\sqrt{x} - 2}{\sqrt{x}+2}\right| < 4\epsilon$$ $$-4\epsilon < \frac{\sqrt{x} - 2}{\sqrt{x}+2} < 4\epsilon$$
First inequation: $$-4\epsilon < \frac{\sqrt{x} - 2}{\sqrt{x}+2}$$ Multiplying by $\sqrt{x} + 2$ (which is positive): $$-4\epsilon (\sqrt{x}+2) < \sqrt{x} - 2$$ $$-4\epsilon \sqrt{x} -8\epsilon < \sqrt{x} - 2$$ $$ - 8\epsilon + 2 < \sqrt{x} + 4\epsilon \sqrt{x}$$ $$ 2(1 - 4\epsilon) < (1 + 4\epsilon) \sqrt{x}$$ $$ \frac{2(1 - 4\epsilon)}{1+4\epsilon} < \sqrt{x}$$ $$ \left[\frac{2(1 - 4\epsilon)}{1+4\epsilon}\right]^2 < (\sqrt{x})^2$$ $$ \frac{4(1 - 4\epsilon)^2}{(1+4\epsilon)^2} < x$$ $$ \frac{4(1 - 4\epsilon)^2}{(1+4\epsilon)^2} - 4 < x - 4$$ $$ 4\left[\frac{(1 - 4\epsilon)^2}{(1+4\epsilon)^2} - 1\right] < x - 4$$ $$ 4\left[\frac{(1 - 4\epsilon)^2 - (1 + 4\epsilon)^2}{(1+4\epsilon)^2}\right] < x - 4$$ $$ 4\left[\frac{1 - 8\epsilon + 16\epsilon^2 - 1 - 8\epsilon - 16\epsilon^2}{(1+4\epsilon)^2}\right] < x - 4$$ $$ 4\left[\frac{-16\epsilon}{(1+4\epsilon)^2}\right] < x - 4$$ $$ \frac{-72\epsilon}{(1+4\epsilon)^2} < x - 4$$
Second inequation: $$\frac{\sqrt{x} - 2}{\sqrt{x}+2} < 4\epsilon$$ Multiplying by $\sqrt{x} + 2$ (which is positive): $$\sqrt{x} - 2 < 4\epsilon (\sqrt{x}+2)$$ $$\sqrt{x} - 2 < 4\epsilon \sqrt{x} + 8\epsilon$$ $$\sqrt{x} - 4\epsilon \sqrt{x} < 8\epsilon + 2$$ $$ (1 - 4\epsilon) \sqrt{x} < 2(1 + 4\epsilon)$$ $$\sqrt{x} < \frac{2(1 + 4\epsilon)}{1-4\epsilon}$$ $$(\sqrt{x})^2 < \left[\frac{2(1 + 4\epsilon)}{1-4\epsilon}\right]^2$$ $$x < \frac{4(1 + 4\epsilon)^2}{(1-4\epsilon)^2}$$ $$x - 4 < \frac{4(1 + 4\epsilon)^2}{(1-4\epsilon)^2} - 4$$ $$x - 4 < 4\left[\frac{(1 + 4\epsilon)^2}{(1-4\epsilon)^2} - 1\right]$$ $$x - 4 < 4\left[\frac{(1 + 4\epsilon)^2 - (1 - 4\epsilon)^2}{(1-4\epsilon)^2}\right]$$ $$x - 4 < 4\left[\frac{1 + 8\epsilon + 16\epsilon^2 - 1 + 8\epsilon - 16\epsilon^2}{(1-4\epsilon)^2}\right]$$ $$x - 4 < 4\left[\frac{16\epsilon}{(1-4\epsilon)^2}\right]$$ $$x - 4 < \frac{72\epsilon}{(1-4\epsilon)^2}$$
Therefore: $$- \frac{72\epsilon}{(1+4\epsilon)^2} < x - 4 < \frac{72\epsilon}{(1-4\epsilon)^2}$$
Taking the minimum of $\frac{72\epsilon}{(1+4\epsilon)^2}$ and $\frac{72\epsilon}{(1-4\epsilon)^2}$ as the value for $\delta$:
$$ \delta = \min\left[\frac{72\epsilon}{(1+4\epsilon)^2} , \frac{72\epsilon}{(1-4\epsilon)^2}\right]$$ $$ \delta = \frac{72\epsilon}{(1+4\epsilon)^2}$$
Second part: Prove
Prove that
$$ 0 < \left|x-4\right| < \frac{72\epsilon}{(1+4\epsilon)^2} \implies \left|\frac{\sqrt{x}-2}{x-4} - \frac{1}{4}\right| < \epsilon$$
Is it correct so far? How to proceed from here?
Correct. Now,
$$0 < \left|x-4\right| < \frac{72\epsilon}{(1+4\epsilon)^2} \implies \frac{-72\epsilon}{(1+4\epsilon)^2} < x-4 < \frac{72\epsilon}{(1+4\epsilon)^2}$$
But: $$\frac{72\epsilon}{(1+4\epsilon)^2} < \frac{72\epsilon}{(1-4\epsilon)^2}$$
Hence:
$$\frac{-72\epsilon}{(1+4\epsilon)^2} < x-4 < \frac{72\epsilon}{(1-4\epsilon)^2}$$
I am sure you can proceed from here.