Why is this limit 1?

1. Oct 13, 2007

frasifrasi

I am looking at the limit as x goes to infinity of x/(x+1)...

If you plug in infinity, won't it produce an indeterminate form inf/inf ? And then by taking the derivative, I get 1/(x+1)^2, which goes to 0... So can anyone explain why the limit is 1 as opposed to 0?

2. Oct 13, 2007

Hurkyl

Staff Emeritus
What does that have to do with anything?

3. Oct 13, 2007

dextercioby

And how do you "plug in infinity" ...?

4. Oct 13, 2007

rocomath

it's

$$\frac{x}{x+1}$$

not

$$\frac{x}{x^{-1}}$$

5. Oct 13, 2007

John1987

Simply use l'Hôpital's rule to compute this limit. This indeed gives a value of 1 as x tends to infinity.

6. Oct 13, 2007

oahsen

since the function behaves like the y=1 line then there is nothing wrong about its derivative to be 0 at infinity.

7. Oct 13, 2007

HallsofIvy

Staff Emeritus
Your reference to the derivative in your first post makes me think you were trying L'Hopital's rule, but incorrectly. L'Hopital's rule says that if f(x) and g(x) both go to 0 (or to infinity)
$$\lim_{x\rightarrow a}= \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$$
: differentiate the numerator and denominator separately.

However, a much simpler way to handle limits at infinity, of rational functions, is to divide both numerator and denominator by the highest power of x- here just divide numerator and denominator by x. $\lim_{x\rightarrow \infty} x$ is difficult but $\lim_{x\rightarrow \infty} 1/x$ is easy!

8. Oct 13, 2007

frasifrasi

Yes, I was doing the l'hopital's rule. I understand that technique of factoring by the highest power and realized that is how the limit was obtained in the book.

But my question is, why doesn't l'hopital apply here? Doesn't the limit of the top and bottom function both go to infinity?

9. Oct 13, 2007

arildno

Technique of "factoring"????
What is f and what is g, and what are their respective derivatives?

10. Oct 13, 2007

frasifrasi

Oh, i see...i was deriving the entire function rather than top and bottom...

11. Oct 13, 2007

ZioX

If f(x)->a<infinity then f'(x) must go to zero. Draw a picture.

EDIT: If it is differentiable...this is not an existence inference...or maybe it is. Let me think.

Last edited: Oct 13, 2007
12. Oct 13, 2007

rock.freak667

$$\frac{x}{x+1}=1-\frac{1}{x+1}$$

$$=1-\frac{\frac{1}{x}}{1+\frac{1}{x}}$$

Now as $x\rightarrow\infty$,$\frac{1}{x}\rightarrow0$

and therefore $$\frac{\frac{1}{x}}{1+\frac{1}{x}}\rightarrow0$$

and thus the limit is 1