Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.4 Limits and Continuity - Exercises - Page 67: 55


$f (x)$ has a removable discontinuity at $x = 4$

Work Step by Step

We find the limit of the given function at $x=4$: \begin{align*} \lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4}&=\lim _{x \rightarrow 4}(x+4)\\ &=8 \end{align*} We see that the function value does not equal the limit at $x=4$: $$ f(4)=10\neq\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4} =8$$ Thus, $f (x)$ has a removable discontinuity at $x = 4$. To remove the discontinuity, we must redefine $f (4) = 8$
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