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 449: 59

Answer

The limit does not exist.

Work Step by Step

Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} \dfrac{x(2+\sin 2x)}{x+1}$ But $\lim\limits_{x \to +\infty} \dfrac{x(2+\sin 2x)}{x+1}$ does not show any indeterminate form . So, we will not apply the L'Hospital's rule . $\lim\limits_{x \to +\infty} \dfrac{x(2+\sin 2x)}{x+1}=\lim\limits_{x \to +\infty} \dfrac{2+\sin 2x)}{1+\dfrac{1}{x}}$ We can see that there is no finite value . So, the limit does not exist.
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