Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 17

Answer

$$\lim\limits_{x\to\infty}(\sqrt{x^2+4x+1}-x)=2$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}(\sqrt{x^2+4x+1}-x)$$ Multiply the function by $\frac{\sqrt{x^2+4x+1}+x}{\sqrt{x^2+4x+1}+x}$, we have: $$A=\lim\limits_{x\to\infty}\frac{(\sqrt{x^2+4x+1}-x)(\sqrt{x^2+4x+1}+x)}{\sqrt{x^2+4x+1}+x}$$ - Consider the numerator: $(\sqrt{x^2+4x+1}-x)(\sqrt{x^2+4x+1}+x)=(x^2+4x+1)-x^2=4x+1$ So, $$A=\lim\limits_{x\to\infty}\frac{4x+1}{\sqrt{x^2+4x+1}+x}$$ Divide both numerator and denominator by $x$, we have - Numerator: $$\frac{4x+1}{x}=4+\frac{1}{x}$$ - Denominator: As $x\to\infty$, we only care for the values of $x\gt0$, which means $\sqrt{x^2}=|x|=x$ So, $$\frac{\sqrt{x^2+4x+1}+x}{x}=\frac{\sqrt{x^2+4x+1}}{x}+1=\sqrt{\frac{x^2+4x+1}{x^2}}+1=\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+1$$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{4+\frac{1}{x}}{\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+1}$$$$A=\frac{4+0}{\sqrt{1+4\times0+0}+1}$$$$A=2$$
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