Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 73: 47


$$\lim _{t \rightarrow-1}(4 t-2 a t+3 a) =5a-4$$

Work Step by Step

Given $$\lim _{t \rightarrow-1}(4 t-2 a t+3 a), \ \ \ a \ \ \text{is constant}$$ let $$ f(t) = (4 t-2 a t+3 a)$$ Since, we have $$ f(-1)= (-4+2 a+3 a)=5a-4$$ So, we get \begin{aligned} L&=\lim _{t \rightarrow-1}(4 t-2 a t+3 a)\\ &= (-4+2 a+3 a)\\ &=5a-4 \end{aligned}
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.