Prove that the limit of $\lim\limits_{x \to 4}{\frac{\sqrt{x}-2}{x-4}} = \frac{1}{4}$

calculus limits epsilon-delta

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.

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