Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises - Page 77: 38

Answer

See proof below.

Work Step by Step

If $a \gt 0$, note that $|a|=a.$ When x approaches a, being very close to $a$, x is positive, so $|x|=x.$ So, when x approaches a,$ |x|$=$x$ approaches $a$ , which is, in this case, equal to $|a|$. In other words, for $a \gt 0$, $\displaystyle \lim_{x\rightarrow a}|x|=|a|$ If $a \lt 0$, note that $|a|=-a.$ when x approaches a, being very close to $a$, x is negative, so $|x|=-x.$ So, when x approaches a, $|x|=-x$ (which is now positive) approaches $-a$ (also positive), which is, in this case, equal to $|a|$. In other words, for $a \lt 0$, $\displaystyle \lim_{x\rightarrow a}|x|=|a|$ For the case $a=0$, the last exercise showed $\displaystyle \lim_{x\rightarrow 0}|x|=0=|0|.$ Thus, whatever the value of a, it is always that $\displaystyle \lim_{x\rightarrow a}|x|=|a|$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.