Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 38

Answer

$$\lim\limits_{x\to\infty}\frac{\sin^2x}{x^2+1}=0$$

Work Step by Step

$$\lim\limits_{x\to\infty}\frac{\sin^2 x}{x^2+1}$$ *Strategy: Apply the Squeeze Theorem 1) We see that $$0\le \sin^2x\le1$$ Also, since $x^2\ge0$ for $x\in R$, therefore $x^2+1\gt0$ for $x\in R$ So, $$\frac{0}{x^2+1}\le\frac{\sin^2x}{x^2+1}\le\frac{1}{x^2+1}$$ (the inequality direction remains because $x^2+1\gt0$) $$0\le\frac{\sin^2x}{x^2+1}\le\frac{1}{x^2+1}\hspace{.5cm}(1)$$ 2) We calculate: - $\lim\limits_{x\to\infty}0=0$ -$\lim\limits_{x\to\infty}\frac{1}{x^2+1}=\frac{1}{\lim\limits_{x\to\infty}(x^2)+1}=\frac{1}{\infty+1}=0$ So, $\lim\limits_{x\to\infty}0=\lim\limits_{x\to\infty}\frac{1}{x^2+1}=0\hspace{.5cm}(2)$ 3) From $(1)$ and $(2)$, according to the Squeeze Theorem, we conclude: $$\lim\limits_{x\to\infty}\frac{\sin^2x}{x^2+1}=0$$
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