Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.5 L'Hopital's Rule; Indeterminate Forms - Exercises Set 6.5 - Page 448: 23

Answer

$-\dfrac{5}{3}$

Work Step by Step

Our aim is to evaluate the limit for $\lim\limits_{x \to -\dfrac{\pi}{2}}\sec(3x) \cos (5x)=\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\cos 5x}{\cos 3x}$. But $\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\cos 5x}{\cos 3x}=\dfrac{\infty}{\infty}$ We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$ where, $a$ can be any real number, infinity or negative infinity. $\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\cos 5x}{\cos 3x}=\lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{-5 \sin (5x)}{-3 \sin (3x)} \\ =\dfrac{5}{3} \times \lim\limits_{x \to -\dfrac{\pi}{2}} \dfrac{\sin (5\pi/2)}{\sin (3 \pi/2)} \\=\dfrac{5}{3} \times \dfrac{1}{-1} \\=-\dfrac{5}{3}$
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