Question

Use the limit theorem and the properties of limits to find the horizontal asymptotes of the graph of the function
Picture below

Answer 1

Answer:

$a. y=0$

Step-by-step explanation:

The horizontal asymptote of the function is given by;

$\lim_{x \to \infty} \frac{3x+x^2-4}{2-x^3+x^2}$

We divide the numerator and the denominator by $x^3$ to get

$\lim_{x \to \infty} \frac{\frac{3x}{x^3}+\frac{x^2}{x^3}-\frac{4}{x^3}}{\frac{2}{x^3}-\frac{x^3}{x^3}+\frac{x^2}{x^3}}$

$\lim_{x \to \infty} \frac{\frac{3}{x^2}+\frac{1}{x}-\frac{4}{x^3}}{\frac{2}{x^3}-1+\frac{1}{x}}$

As $x\rightarrow \infty,\frac{1}{x^n}\rightarrow 0$

$\lim_{x \to \infty} \frac{\frac{3}{x^2}+\frac{1}{x}-\frac{4}{x^3}}{\frac{2}{x^3}-1+\frac{1}{x}}=\frac{0+0-0}{0-1+0}=0$

Therefore the horizontal asymptote is $y=0$