Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Review - Exercises: 28

Answer

Using $\varepsilon$ and $\delta$ neighborhoods we show that, indeed, $\lim_{x\to4^+}\frac{2}{\sqrt{x-4}}=\infty$.

Work Step by Step

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 $\left|\frac{2}{\sqrt{x-4}}\right|>\varepsilon.$
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