Question

a. Analyze ﻿lim⁡x→∞f(x){\displaystyle\lim_{x	o\infty}{f(x)}}x→∞lim​f(x)﻿ and﻿lim⁡x→−∞f(x){\displaystyle\lim_{x	o-\infty}{f(x)}}x→−∞lim​f(x)﻿ for each function.f(x)=1+x−2x2−x3x2+1f\left(xight)=\frac{$$1+x-2x^2$-x^3$$}{$$x^2+1$$}

Accepted Answer

Identify the highest degree term in both the numerator and the denominator. In the given function $$ f(x) = \frac{1 + x - 2x^2 - x^3}{x^2 + 1} $$, the highest degree term in the numerator is $$-x^3$$ and in the denominator is $$x^2. $$Divide every term in the numerator and the denominator by $$x^2$$, the highest degree term in the denominator, to simplify the expression. This gives $$ f(x) = \frac{\frac{1}{x^2} + \frac{x}{x^2} - \frac{2x^2}{x^2} - \frac{x^3}{x^2}}{\frac{x^2}{x^2} + \frac{1}{x^2}} . $$Simplify the expression: $$ f(x) = \frac{\frac{1}{x^2} + \frac{1}{x} - 2 - x}{1 + \frac{1}{x^2}} . $$Evaluate the limit as $$ x 	o \infty . As$$ $$ x 	o \infty $$, the terms $$ \frac{1}{x^2} $$ and $$ \frac{1}{x} $$ approach 0. Thus, the expression simplifies to $$ \frac{0 + 0 - 2 - x}{1 + 0} = -x - 2 . $$Therefore, $$ \lim_{x 	o \infty} f(x) = -\infty . $$Evaluate the limit as $$ x 	o -\infty . $$Similarly, as $$ x 	o -\infty $$, the terms $$ \frac{1}{x^2} $$ and $$ \frac{1}{x} $$ approach 0. The expression simplifies to $$ \frac{0 + 0 - 2 - x}{1 + 0} = -x - 2 . $$Therefore, $$ \lim_{x 	o -\infty} f(x) = \infty .$$

a. Analyze ${\(\displaystyle\[\lim\)_{x\(\to\]\infty\)}{f(x)}}$ and ${\(\displaystyle\)\(\lim\)_{x\(\to\)-\(\infty\)}{f(x)}}$ for each function.

$f (x) = \frac{1 + x - 2 x^{2} - x^{3}}{x^{2} + 1}$

Verified video answer for a similar problem:

Key Concepts

Limits at Infinity

Rational Functions

Dominant Terms

a. Analyze limx→∞f(x){\(\displaystyle\[\lim\)_{x\(\to\]\infty\)}{f(x)}}x→∞limf(x) andlimx→−∞f(x){\(\displaystyle\)\(\lim\)_{x\(\to\)-\(\infty\)}{f(x)}}x→−∞limf(x) for each function.f(x)=1+x−2x2−x3x2+1f\(\left\)(x\(\right\))=\(\frac{1+x-2x^2-x^3}{x^2+1}\)