Answer
$$\lim_{x\to a}\frac{x^2-a^2}{x^4-a^4}=\frac{1}{2a^2}$$
Work Step by Step
$$A=\lim_{x\to a}\frac{x^2-a^2}{x^4-a^4}=\lim_{x\to a}\frac{x^2-a^2}{(x^2-a^2)(x^2+a^2)}=\lim_{x\to a}\frac{1}{x^2+a^2}$$
$$A=\frac{1}{a^2+a^2}=\frac{1}{2a^2}$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 2 - Practice Exercises - Page 111: 12
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