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 78: 68


$b=-6$ $\lim\limits_{x \to 2} f(x)=0$

Work Step by Step

We are given the function: $f(x)=\begin{cases} 3x+b,\text{ if }x\leq 2\\ x-2,\text{ if }x>2 \end{cases}$ For $\lim\limits_{x \to 2}$ to exist we must have: $\lim\limits_{x \to 2^-} f(x)=\lim\limits_{x \to 2^+} f(x)$ (1) Determine the left hand limit and the right hand limit: $\lim\limits_{x \to 2^-} f(x)=\lim\limits_{x \to 2^-} (3x+b)=3(2)+b=6+b$ $\lim\limits_{x \to 2^+} f(x)=\lim\limits_{x \to 2^+} (x-2)=2-2=0$ For the condition (1) to be checked, we must have: $6+b=0\Rightarrow b=-6$ The value of the limit is: $\lim\limits_{x \to 2} f(x)=0$
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.