# Thread: Asymptotes of a function

1. ## Asymptotes of a function

y = (2x)/(x+1) When solving for the horizontal asymptote and substituting infinity in for x, infinity just crosses out and i'm left with 2/1 making the horizontal asymptote y = 2. right?

2. Originally Posted by VitaX
y = (2x)/(x+1) When solving for the horizontal asymptote and substituting infinity in for x, infinity just crosses out and i'm left with 2/1 making the horizontal asymptote y = 2. right?
You cannot substitute infinity. It's not a number.

Approach 1:

$y = \frac{2x}{x + 1} = \frac{2}{1 + \frac{1}{x}}$.

$\lim_{x \rightarrow \pm \infty} \frac{2}{1 + \frac{1}{x}} = 2$.

Therefore $\lim_{x \rightarrow \pm \infty} y = 2$ and so $y = 2$ is the horizontal asymptote.

Approach 2:

$y = \frac{2x}{x + 1} = 2 \left( \frac{x}{x + 1} \right) = 2 \left( 1 - \frac{1}{x + 1}\right) = 2 - \frac{2}{x + 1}$

and by inspection the horizontal asymptote is $y = 2$.

Other approaches are possible.

3. Originally Posted by mr fantastic
You cannot substitute infinity. It's not a number.

Approach 1:

$y = \frac{2x}{x + 1} = \frac{2}{1 + \frac{1}{x}}$.

$\lim_{x \rightarrow \pm \infty} \frac{2}{1 + \frac{1}{x}} = 2$.

Therefore $\lim_{x \rightarrow \pm \infty} y = 2$ and so $y = 2$ is the horizontal asymptote.

Approach 2:

$y = \frac{2x}{x + 1} = 2 \left( \frac{x}{x + 1} \right) = 2 \left( 1 - \frac{1}{x + 1}\right) = 2 - \frac{2}{x + 1}$

and by inspection the horizontal asymptote is $y = 2$.

Other approaches are possible.
You say you cannot substitute infinity because it is not a number. But my teacher has said on several occassions it is a number but not a measurable one. But when infinity is in the denominator of any equation example : 6/infinity that is 0. So perhaps I just understood you wrong.

4. Originally Posted by VitaX
You say you cannot substitute infinity because it is not a number. But my teacher has said on several occassions it is a number but not a measurable one. But when infinity is in the denominator of any equation example : 6/infinity that is 0. So perhaps I just understood you wrong.
No, you have understood me perfectly. Infinity is not a number and cannot be substituted into an expression. To do so is sloppy and mathematically wrong.

When dealing with infinity, you have to take a limit. 6/infinity = 0 is a complete nonsense, $\lim_{x \rightarrow + \infty} \frac{6}{x} = 0$ is the correct way of treating it.

5. Originally Posted by mr fantastic
No, you have understood me perfectly. Infinity is not a number and cannot be substituted into an expression. To do so is sloppy and mathematically wrong.

When dealing with infinity, you have to take a limit. 6/infinity = 0 is a complete nonsense, $\lim_{x \rightarrow + \infty} \frac{6}{x} = 0$ is the correct way of treating it.
Yes i understand I just didn't say take the limit of 1/x as x approaches infinity. I understand the terminology, just not familiar with latex.