Answer
$3$
Work Step by Step
Here, we have $f(x)=3x-4;x=2$
and
$f(x+h)=3(x+h)-4=3x+3h-4$
Therefore, $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\dfrac{(3x+3h-4)-(3x-4)}{h}$
$\implies \lim_{h\to0}\dfrac{3x+3h-4-3x+4}{h}=3$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 59
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