Thread: 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?

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,

Divide by
Thus,

So basically, the limit,

One of standard limits.
Thus, you have division by zero with non-zero numerator thus the limit is infinite. (does not exist).

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

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.

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!