University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 95: 32


$$\lim_{t\to0}\sin(\frac{\pi}{2}\cos(\tan t))=1$$ The function is continuous at $t=0$

Work Step by Step

*Recall Theorem 10: If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$ $$A=\lim_{t\to0}f(t)=\lim_{t\to0}\sin(\frac{\pi}{2}\cos(\tan t))$$ For the domain $(-\infty,\infty)$, $\lim_{x\to c}\sin x= \sin c$ and $\lim_{x\to c}\cos x= \cos c$ So both functions $y=\sin x$ and $y=\cos x$ are continuous on $(-\infty,\infty)$ That means we can apply Theorem 10 here. In detail, $$A=\sin(\frac{\pi}{2}\lim_{t\to0}\cos(\tan t))$$ $$A=\sin(\frac{\pi}{2}\cos(\lim_{t\to0}\tan t))$$ $$A=\sin(\frac{\pi}{2}\cos(\tan0))$$ $$A=\sin(\frac{\pi}{2}\cos0)$$ $$A=\sin(\frac{\pi}{2}\times1)$$ $$A=\sin\frac{\pi}{2}=1$$ - Examine $f(0)$: $$\sin(\frac{\pi}{2}\cos(\tan0))=\sin(\frac{\pi}{2}\cos0)=\sin(\frac{\pi}{2}\times1)=\sin\frac{\pi}{2}=1$$ As $\lim_{t\to0}f(t)=f(0)$, the function is continuous at $t=0$
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