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?

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- Sep 19th 2009, 11:31 PMVitaXAsymptotes 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?

- Sep 20th 2009, 12:18 AMmr fantastic
You cannot substitute infinity. It's not a number.

**Approach 1:**

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

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

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

**Approach 2:**

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

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

Other approaches are possible. - Sep 20th 2009, 12:42 AMVitaX
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.

- Sep 20th 2009, 12:52 AMmr 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, $\displaystyle \lim_{x \rightarrow + \infty} \frac{6}{x} = 0$ is the correct way of treating it. - Sep 20th 2009, 12:54 AMVitaX