University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 107: 11

Answer

$$\lim_{t\to-\infty}\frac{2-t+\sin t}{t+\cos t}=1$$

Work Step by Step

$$A=\lim_{t\to-\infty}\frac{2-t+\sin t}{t+\cos t}$$ Here we can divide both numerator and denominator by $t$, the highest power of $t$ in the denominator: $$A=\lim_{t\to-\infty}\frac{\frac{2}{t}-1+\frac{\sin t}{t}}{1+\frac{\cos t}{t}}$$ - We already know $\lim_{t\to-\infty}(2/t)=0$ - For $\lim_{t\to-\infty}(\sin t)/t$: As $-1\le\sin t\le1$, we have $(-1/t)\le(\sin t/t)\le(1/t)$. But we know that $\lim_{t\to-\infty}(-1/t)=\lim_{t\to-\infty}(1/t)=0$ Therefore, following Squeeze Theorem: $\lim_{t\to-\infty}(\sin t)/t=0$ - For $\lim_{t\to-\infty}(\cos t)/t$: As $-1\le\cos t\le1$, we have $(-1/t)\le(\cos t/t)\le(1/t)$. But we know that $\lim_{t\to-\infty}(-1/t)=\lim_{t\to-\infty}(1/t)=0$ Therefore, following Squeeze Theorem: $\lim_{t\to-\infty}(\cos t)/t=0$ Therefore, $$A=\frac{0-1+0}{1+0}=1$$
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