University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 85: 46

Answer

$\lim_{x\to-2^+}f(x)=\lim_{x\to2^-}f(x)=7$ $\lim_{x\to-2^-}f(x)$ cannot be decided here.
1535798208

Work Step by Step

If $f$ is an even function of $x$, then $f(-x)=f(x)$. We know that $\lim_{x\to2^-}f(x)=7$. This means as $x$ approaches $2$ from the left $(x\lt2)$, the values of $f(x)$ gets closer and closer to $7$. $f$ is an even function, so its graph would be symmetric through $Oy$. All the values of $f(x)$ to the left of $2$ would be reflected in the values of $f(x)$ to the right of $-2$. So if $x$ approaches $-2$ from the right, as $-x$, $f(-x)=f(x)$, the values of $f(x)$ would also approach $7$. In other words, $\lim_{x\to-2^+}f(x)=\lim_{x\to2^-}f(x)=7$ The value of $\lim_{x\to-2^-}f(x)$, unfortunately, cannot be decided without the knowledge of $\lim_{x\to2^+}f(x)$. A graph of an even function is shown below for clarification. As you can see, $\lim_{x\to-2^+}f(x)=\lim_{x\to2^-}f(x)=7$ here, but $\lim_{x\to-2^-}f(x)$ would not exist here because the graph is not defined in the domain $(-\infty,-2)$.
Small 1535798208
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.