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

`a. y=0`

Explanation:

The horizontal asymptote of the function is given by;

`\lim_(x \to \infty) (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) ((3x)/(x^3)+(x^2)/(x^3)-(4)/(x^3))/((2)/(x^3)-(x^3)/(x^3)+(x^2)/(x^3))`

`\lim_(x \to \infty) ((3)/(x^2)+(1)/(x)-(4)/(x^3))/((2)/(x^3)-1+(1)/(x))`

As
`x\rightarrow \infty,(1)/(x^n)\rightarrow 0`

`\lim_(x \to \infty) ((3)/(x^2)+(1)/(x)-(4)/(x^3))/((2)/(x^3)-1+(1)/(x))=(0+0-0)/(0-1+0)=0`

Therefore the horizontal asymptote is
`y=0`