Answer
$$1$$
Work Step by Step
Given $$f(x) = \frac{ x^3-1 }{x-1}$$
Then
\begin{align*}
\lim_{x\to 1} f(x) &= \lim_{x\to 1} \frac{ x^3-1 }{x-1}\\
&=\lim_{x\to 1} \frac{ (x-1)(x^2+x+1) }{x-1}\\
&=\lim_{x\to 1}(x^2+x+1)\\
&=3
\end{align*}
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 70: 39
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