University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 65

Answer

(a) $$\lim_{x\to0}\frac{x\sin x}{2-2\cos x}=1$$ (b) As $x\to0$, all of the three graphs approach $y=1$, squeezing together as they go.
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Work Step by Step

(a) - Calculate $\lim_{x\to0}(1-\frac{x^2}{6})$ and $\lim_{x\to0}(1)$ $\lim_{x\to0}\Big(1-\frac{x^2}{6}\Big)=1-\frac{0^2}{6}=1-0=1$ $\lim_{x\to0}1=1$ So, $\lim_{x\to0}(1-\frac{x^2}{6})=\lim_{x\to0}1=1$ - Yet $1-\frac{x^2}{6}\le \frac{x\sin x}{2-2\cos x}\le 1$ for all $x$ close to $0$ Therefore, applying the Sandwich Theorem, we can conclude that $$\lim_{x\to0}\frac{x\sin x}{2-2\cos x}=1$$ A note here is that the inequality does not have to hold for $x=0$ though, if it already holds for all values of $x$ close to $0$ for the simple reason that $\lim_{x\to0}$ accounts for only the values of $x$ close to $0$ and does not account for $0$. (b) The graph is shown below. As $x\to0$, all of the three graphs approach $y=1$, squeezing together as they go.
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