Answer
$$\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=2x$$
Work Step by Step
$$A=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}$$
$$A=\lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h}$$
$$A=\lim_{h\to0}\frac{2xh+h^2}{h}$$
$$A=\lim_{h\to0}(2x+h)$$
$$A=2x+0=2x$$
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: 13
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