Question

1. find the horizontal asymptote(s) of

(i) ( y=\frac{3 x^{9}+5 x+2}{4 x^{9}-7 x^{4}+3 x} ) (ii) ( y=\frac{4 e^{2 x}+3 e^{x}+5}{left(1-2 e^{x}
ight)left(e^{x}+3
ight)} ) (iii) ( y=\frac{sqrt{4 x^{2}+7}}{5 x-3} ).

Explanation:

Step1: Find the horizontal asymptote for \(y = \frac{3x^{9}+5x + 2}{4x^{9}-7x^{4}+3x}\)

For a rational function \(y=\frac{f(x)}{g(x)}\) where \(f(x)=a_nx^n+\cdots+a_0\) and \(g(x)=b_mx^m+\cdots + b_0\), if \(n = m\), the horizontal asymptote is \(y=\frac{a_n}{b_m}\). Here \(n = m=9\), \(a_n = 3\), \(b_m=4\). So \(y=\frac{3}{4}\)

Step2: Find the horizontal asymptote for \(y=\frac{4e^{2x}+3e^{x}+5}{(1 - 2e^{x})(e^{x}+3)}\)

First, expand the denominator \((1 - 2e^{x})(e^{x}+3)=e^{x}+3-2e^{2x}-6e^{x}=-2e^{2x}-5e^{x}+3\)
Divide numerator and denominator by \(e^{2x}\):

$$
LATEXBLOCK0
$$

As \(x
ightarrow\infty\), \(\frac{1}{e^{x}}
ightarrow0\) and \(\frac{1}{e^{2x}}
ightarrow0\). So \(y = - 2\)

Step3: Find the horizontal asymptote for \(y=\frac{\sqrt{4x^{2}+7}}{5x - 3}\)

Consider two cases: \(x
ightarrow\infty\) and \(x
ightarrow-\infty\)
For \(x
ightarrow\infty\):

$$
LATEXBLOCK1
$$

For \(x
ightarrow-\infty\):

$$
LATEXBLOCK2
$$

Answer:

(i) \(y = \frac{3}{4}\)
(ii) \(y=-2\)
(iii) \(y=\frac{2}{5}\) and \(y =-\frac{2}{5}\)