Question

find the end - behavior of the following rational functions. $g(x)=\frac{x + 1}{(2x + 5)(x - 4)}$ $lim_{x
ightarrow-infty}f(x)=0,lim_{x
ightarrowinfty}f(x)=\frac{1}{2}$ $lim_{x
ightarrow-infty}f(x)=\frac{1}{2},lim_{x
ightarrowinfty}f(x)=0$ $lim_{x
ightarrow-infty}f(x)=0,lim_{x
ightarrowinfty}f(x)=0$ $lim_{x
ightarrow-infty}f(x)=1,lim_{x
ightarrowinfty}f(x)=1$

Explanation:

Step1: Expand the numerator

First, expand \((2x + 5)(x - 4)\) using the FOIL - method. \((2x+5)(x - 4)=2x\times x-2x\times4 + 5\times x-5\times4=2x^{2}-8x + 5x-20=2x^{2}-3x - 20\). So, \(g(x)=\frac{2x^{2}-3x - 20}{x + 1}\).

Step2: Use long - division or the fact that for a rational function \(y=\frac{f(x)}{d(x)}\) where \(f(x)=a_nx^n+\cdots+a_0\) and \(d(x)=b_mx^m+\cdots+b_0\) (\(n\geq m\)), we can analyze the end - behavior by considering the leading terms.

The degree of the numerator \(n = 2\) and the degree of the denominator \(m = 1\). When \(x\to\pm\infty\), we consider the ratio of the leading terms of the numerator and the denominator. The leading term of the numerator is \(2x^{2}\) and the leading term of the denominator is \(x\). \(\lim_{x\to\pm\infty}\frac{2x^{2}-3x - 20}{x + 1}=\lim_{x\to\pm\infty}\frac{2x^{2}}{x}=\lim_{x\to\pm\infty}2x=\pm\infty\).
We can also use long - division: \(2x^{2}-3x - 20=(x + 1)(2x-5)-15\), so \(g(x)=2x-5-\frac{15}{x + 1}\). As \(x\to\pm\infty\), \(\lim_{x\to\pm\infty}g(x)=\lim_{x\to\pm\infty}(2x-5-\frac{15}{x + 1})\). Since \(\lim_{x\to\pm\infty}\frac{15}{x + 1}=0\), \(\lim_{x\to-\infty}g(x)=-\infty\) and \(\lim_{x\to\infty}g(x)=\infty\).

Answer:

\(\lim_{x\to-\infty}g(x)=-\infty,\lim_{x\to\infty}g(x)=\infty\)