Question

use the limit theorem and the properties of limits to find the horizontal asymptotes of the graph of the function f(x)= 3x^2+5 / 4x^2-6x+2​

Answer 1

Answer:

$y=\frac{3}{4}$

Step-by-step explanation:

Given: $f(x)=\frac{3x^2+5}{4x^2-6x+2}$

To find:  horizontal asymptotes of the graph of the function

Solution:

Line y = L is a horizontal asymptote of the function y = f(x) if either $\lim_{x\rightarrow \infty }f(x)=L\,,\,\lim_{x\rightarrow \infty^+ }f(x)=L$, and L is finite

$f(x)=\frac{3x^2+5}{4x^2-6x+2}$

Use properties:

$\lim_{x\rightarrow a}(f+g)=\lim_{x\rightarrow a}f+\lim_{x\rightarrow a}g\\\lim_{x\rightarrow a}\left ( \frac{f}{g} \right )=\frac{\lim_{x\rightarrow a}f}{\lim_{x\rightarrow a}g}$

Divide numerator and denominator by $x^2$

$f(x)=\frac{\frac{3x^2+5}{x^2}}{\frac{4x^2-6x+2}{x^2}}\\=\frac{3+\frac{5}{x^2}}{4-\frac{6}{x}+\frac{2}{x^2}}$

Therefore,

$\lim_{x\rightarrow \infty^- }\left [ \frac{3+\frac{5}{x^2}}{4-\frac{6}{x}+\frac{2}{x^2}} \right ]$

$=\frac{\lim_{x\rightarrow \infty^- } \left [ 3+\frac{5}{x^2} \right ]}{\lim_{x\rightarrow \infty^- }\left [ 4-\frac{6}{x}+\frac{2}{x^2} \right ]}\\=\frac{3+0}{4-0+0}\\=\frac{3}{4}$

Also,

$\lim_{x\rightarrow \infty^+ }\left [ \frac{3+\frac{5}{x^2}}{4-\frac{6}{x}+\frac{2}{x^2}} \right ]=\frac{\lim_{x\rightarrow \infty^+ } \left [ 3+\frac{5}{x^2} \right ]}{\lim_{x\rightarrow \infty^+ }\left [ 4-\frac{6}{x}+\frac{2}{x^2} \right ]}\\=\frac{3+0}{4-0+0}\\=\frac{3}{4}$

Here, $\lim_{x\rightarrow \infty^- }\left [ \frac{3+\frac{5}{x^2}}{4-\frac{6}{x}+\frac{2}{x^2}} \right ]=\lim_{x\rightarrow \infty^+ }\left [ \frac{3+\frac{5}{x^2}}{4-\frac{6}{x}+\frac{2}{x^2}} \right ]=\frac{3}{4}$

So, $y=\frac{3}{4}$ is the horizontal asymptote