## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 83

#### Answer

$$\lim_{x\to-\infty}(2x+\sqrt{4x^2+3x-2})=-\frac{3}{4}$$

#### Work Step by Step

$$A= \lim_{x\to-\infty}(2x+\sqrt{4x^2+3x-2})$$ We cannot examine the behavior of $(2x+\sqrt{4x^2+3x-2})$ as $x$ approaches $-\infty$ right away, because it will lead to the unsolved situation of $(-\infty)+\sqrt{\infty}=\infty-\infty$, which we would try to avoid. Instead, we would want to turn it into a rational function and apply the usual method, by doing the followings: $$A=\lim_{x\to-\infty}\Big[(2x+\sqrt{4x^2+3x-2})\times\frac{(2x-\sqrt{4x^2+3x-2})}{(2x-\sqrt{4x^2+3x-2})}\Big]$$ $$A=\lim_{x\to-\infty}\frac{4x^2-(4x^2+3x-2)}{(2x-\sqrt{4x^2+3x-2})}=\lim_{x\to-\infty}\frac{-3x+2}{2x-\sqrt{4x^2+3x-2}}$$ Now we can divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$: $$A=\lim_{x\to-\infty}\frac{-3+\frac{2}{x}}{2-\frac{\sqrt{4x^2+3x-2}}{x}}$$ However, we must be careful with the sign here when we try to put $x$ inside the square root: - We know that $|x|=\sqrt{x^2}$. Here, since $x\to-\infty$, we consider values of $x\lt0$, therefore, $x = -\sqrt{x^2}$. $$A=\lim_{x\to-\infty}\frac{-3+\frac{2}{x}}{2-\frac{\sqrt{4x^2+3x-2}}{-\sqrt{x^2}}}=\lim_{x\to-\infty}\frac{-3+\frac{2}{x}}{2+\sqrt{4+\frac{3}{x}-\frac{2}{x^2}}}$$ $$A=\frac{-3+0}{2+\sqrt{4+0-0}}=\frac{-3}{2+\sqrt4}=\frac{-3}{2+2}=-\frac{3}{4}$$

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