# Limit Question

• November 7th 2006, 06:13 PM
Ideasman
Limit Question
I'm confused:

Why is the limit(as x -> 0) sin(x)/(x-sin(x)) = infinity?

I have done this 10 times now, and I get that it equals -1.

We have an indeterminant form when we plug 0 in (0/0). Thus, I can use l'hopitals rule.

Then:

cos(x)/(1-cos(x)) = 1/0...doesn't work, so do l'hopitals again.

-sin(x)/sin(x) = 0/0; use l'hopitals again

-cos(x)/cos(x) = -1/1 = -1

So, how does it equal infinity?
• November 7th 2006, 07:38 PM
ThePerfectHacker
Quote:

Originally Posted by Ideasman
I'm confused:

Why is the limit(as x -> 0) sin(x)/(x-sin(x)) = infinity?

I have done this 10 times now, and I get that it equals -1.

We have an indeterminant form when we plug 0 in (0/0). Thus, I can use l'hopitals rule.

Then:

cos(x)/(1-cos(x)) = 1/0...doesn't work, so do l'hopitals again.

-sin(x)/sin(x) = 0/0; use l'hopitals again

-cos(x)/cos(x) = -1/1 = -1

So, how does it equal infinity?

Do not use L'Hopitals rule,
You have,
$\frac{\sin x}{x-\sin x}$
Divide by $\sin x$
Thus,
$\frac{1}{\frac{x}{\sin x}-1}$
So basically, the limit,
$\lim_{x\to 0}\frac{x}{\sin x}=1$
One of standard limits.
Thus, you have division by zero with non-zero numerator thus the limit is infinite. (does not exist).
• November 7th 2006, 07:46 PM
Ideasman
Why can't you use l'hopitals rule? It is an indeterminant form, 0/0.
• November 7th 2006, 07:49 PM
Soltras
Quote:

Originally Posted by Ideasman
Why can't you use l'hopitals rule? It is an indeterminant form, 0/0.

You can, but you used it a second time when the form was 1/0, which was incorrect. Arriving at 1/0 actually would imply the answer is infinity.
• November 7th 2006, 07:50 PM
ThePerfectHacker
Quote:

Originally Posted by Ideasman
Why can't you use l'hopitals rule? It is an indeterminant form, 0/0.

You can, but I rather not.

You mistake is that after you use the L-Hopital rule once you have 1/0 which is limits of infinite. YOU CANNOT USE RULE HERE. Its proof is based on the fact that both are zero!