Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 22

Show that the statement $\displaystyle \lim \limits_{x \to 4^+} \frac{2}{\sqrt{x - 4}} = \infty$ using the precise definition of a limit.

Based from the definition, we let $M > 0$. So,

$\qquad$ if $0 < | x - 4 | < \delta$ then $\displaystyle \frac{2}{\sqrt{x - 4}} > M$

But,

$\qquad$ $\displaystyle \frac{2}{\sqrt{x - 4}} > M \to \sqrt{x -4} < \frac{2}{M} \to x - 4 < \frac{(2)^2}{M^2} \to x - 4 < \frac{4}{M^2}$

It shows that if we choose $\delta = \displaystyle \frac{4}{M^2}$, we will be able to prove that

$\qquad$ $\lim \limits_{x \to 4^+} \displaystyle \frac{2}{\sqrt{x - 4}} = \infty$