## Calculus: Early Transcendentals 8th Edition

Using $\varepsilon$ and $\delta$ neighborhoods we show that, indeed, $\lim_{x\to4^+}\frac{2}{\sqrt{x-4}}=\infty$.
The precise definition of $$\lim_{x\to4^+}\frac{2}{\sqrt{x-4}}=+\infty$$ is: For every $\varepsilon>0$ there is $\delta>0$ such that when $0\varepsilon$. It is enough to take $\delta=\frac{4}{\varepsilon^2}$ because then $$\left|\frac{2}{\sqrt{x-4}}\right|=\frac{2}{\sqrt{|x-4|}}>\frac{2}{\sqrt{\delta}}=\frac{2}{\sqrt{\frac{4}{\varepsilon^2}}}=\frac{2\varepsilon}{2}=\varepsilon.$$ So, indeed, for every $\varepsilon>0$ we can choose $\delta=\frac{4}{\varepsilon^2}$ to have $\left|\frac{2}{\sqrt{x-4}}\right|>\varepsilon.$