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 94: 23


$y=\frac{x\tan x}{x^2+1}$ is continuous for all $x\ne\pi/2+k\pi$

Work Step by Step

$$y=\frac{x\tan x}{x^2+1}=\frac{x\times \frac{\sin x}{\cos x}}{x^2+1}=\frac{x\sin x}{\cos x(x^2+1)}$$ - Domain: $y$ is defined where $\cos x\ne0$ and $x^2+1\ne0$: For $\cos x\ne0$, then $x\ne\pi/2+k\pi$ $(k\in Z)$ For $x^2+1\ne0$, since $x^2\ge0$, it leads to $x^2+1\gt0$ for all $x\in R$ So our domain here is all $x\in R$ except where $x=\pi/2+k\pi$ $(k\in Z)$. - As $x$ approaches any values of $\pi/2+k\pi$, $\cos x$ would approach $0$, thus $\frac{x\sin x}{\cos x(x^2+1)}$ will approach infinity and does not reach for any definite value. In other words, $\lim_{x\to(\pi/2+k\pi)}\frac{x\sin x}{\cos x(x^2+1)}$ does not exist, so the function $y=\frac{x\tan x}{x^2+1}$ is discontinuous at all points $x=\pi/2+k\pi$ $(k\in Z)$. So for all $x\ne\pi/2+k\pi$: - $\lim_{x\to c}\frac{x}{x^2+1}=\frac{c}{c^2+1}$. So $y=\frac{x}{x^2+1}$ is continuous in the domain defined. - $\lim_{x\to c}\tan x=\tan c$. So $y=\tan x$ is continuous in the domain defined. According to Theorem 8, the product of two functions continuous at $x=c$ is also continuous at $x=c$. Therefore, $y=\frac{x\tan x}{x^2+1}$ is continuous for all $x\ne\pi/2+k\pi$ $(k\in Z)$
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