Find each limit, if it exists. 2x2 + xlim6x + 2x2 + x-00

Given, limx→-∞ 2x2+x6x3+2x2+x

Answered: 2x2 + x lim 6x + 2x2 + x -00

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter4: Polynomials

Section4.5: Multiplying Polynomials By Monimials

Problem 6OE

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Question

Find each limit, if it exists.

Expert Solution

Step 1

Given,

$\lim_{x \to - \infty} \frac{2 x^{2} + x}{6 x^{3} + 2 x^{2} + x}$

Step 2

Now the limit

$\begin{array}{rcl} \lim_{x \to - \infty} \frac{2 x^{2} + x}{6 x^{3} + 2 x^{2} + x} & = & \lim_{x \to - \infty} \frac{x^{3} (\frac{2}{x} + \frac{1}{x^{2}})}{x^{3} (6 + \frac{2}{x} + \frac{1}{x^{2}})} \\ = & \lim_{x \to - \infty} \frac{(\frac{2}{x} + \frac{1}{x^{2}})}{(6 + \frac{2}{x} + \frac{1}{x^{2}})} \\ = & \frac{(\frac{2}{\infty} + \frac{1}{\infty})}{(6 + \frac{2}{- \infty} + \frac{1}{\infty})} \\ = & \frac{(0 + 0)}{(6 + 0 + 0)} \\ = & \frac{0}{6} \\ = & 0 \end{array}$

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