# Problem with a limit as infinity

1. Jul 27, 2004

### rumaithya

Hello,

There is a question in Calculus By James Stewart (5e) that says:
Find the horizontal and vertical asympotes of
$$y = \frac{x^2 + 4}{x^2 - 1}}$$

I put it in limit as

$$\lim_{\substack{x\rightarrow \infty}} \frac{x^2 + 4}{x^2 - 1}}$$

and divided the numeratir abd denominator by$$x^2$$

and I got $$\frac{1}{1}} = 1$$ which would be the horizontal asympotes.
And then I stopped here I don't know how to find the vertical asympote and i'm not sure about the horizontal asympote.

The examples in the book doesn't look the same. Each time they use different method to solve them. And now I'm getting confused.

Can someone help ?

Thanks

2. Jul 27, 2004

Horizontal asymptotes: limit as x goes to positive and negative infinity.

Vertical asymptotes: the value(s) a such that limit as x goes to a, f(x) goes to positive or negative infinity. Found by setting the denominator equal to zero.

3. Jul 28, 2004

### Corneo

To find vertical asymptotes, you set the denominator to equal 0. For x values close to where the denominator will equal zero, the function will blow up. For your case,

$$x^2-1=0$$
$$x^2=1$$
$$x=\pm 1$$

Also you are correct to take the limit to determine any horizontal asymptoes.
The answer is y=1.