2. find the horizontal asymptote(s) of\n(i) ( y=\frac{3 x^{9}+5 x+2}{4 x^{9}-7 x^{4}+3 x} ) (ii) ( y=\frac{4…

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}): [ \begin{align*} y&=\lim_{x\rightarrow\infty}\frac{4e^{2x}+3e^{x}+5}{-2e^{2x}-5e^{x}+3}\ &=\lim_{x\rightarrow\infty}\frac{4+\frac{3}{e^{x}}+\frac{5}{e^{2x}}}{-2-\frac{5}{e^{x}}+\frac{3}{e^{2x}}}\ \end{align*} ] As (x\rightarrow\infty), (\frac{1}{e^{x}}\rightarrow0) and (\frac{1}{e^{2x}}\rightarrow0). So (y = - 2)

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

Consider two cases: (x\rightarrow\infty) and (x\rightarrow-\infty) For (x\rightarrow\infty): [ \begin{align*} y&=\lim_{x\rightarrow\infty}\frac{\sqrt{4x^{2}+7}}{5x - 3}\ &=\lim_{x\rightarrow\infty}\frac{\vert x\vert\sqrt{4+\frac{7}{x^{2}}}}{5x - 3}\ &=\lim_{x\rightarrow\infty}\frac{x\sqrt{4+\frac{7}{x^{2}}}}{5x - 3}\ &=\lim_{x\rightarrow\infty}\frac{\sqrt{4+\frac{7}{x^{2}}}}{5-\frac{3}{x}}=\frac{2}{5} \end{align*} ] For (x\rightarrow-\infty): [ \begin{align*} y&=\lim_{x\rightarrow-\infty}\frac{\sqrt{4x^{2}+7}}{5x - 3}\ &=\lim_{x\rightarrow-\infty}\frac{\vert x\vert\sqrt{4+\frac{7}{x^{2}}}}{5x - 3}\ &=\lim_{x\rightarrow-\infty}\frac{-x\sqrt{4+\frac{7}{x^{2}}}}{5x - 3}\ &=\lim_{x\rightarrow-\infty}\frac{-\sqrt{4+\frac{7}{x^{2}}}}{5-\frac{3}{x}}=-\frac{2}{5} \end{align*} ]

Answer:

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