# Trig Limits

1. Nov 5, 2013

### Qube

1. The problem statement, all variables and given/known data

http://i.minus.com/iCJwlfzPc5fRu.png [Broken]

2. Relevant equations

This feels like a movie requiring the suspension of disbelief.

3. The attempt at a solution

cot(x) = cos(x)/sin(x).

cot(0) = 1/0

Right? How in the world then is 0cot(0) - 1 = 0? That should be infinity minus 1.

Last edited by a moderator: May 6, 2017
2. Nov 5, 2013

### Dick

You can't necessarily just substitute 0 in. You have to think about limits. Write the first term in the numerator as 10x*cot(10*x)=10x*cos(10x)/sin(10x). 10x/sin(10x) has a simple limit. What is it?

Last edited by a moderator: May 6, 2017
3. Nov 5, 2013

### Integral

Staff Emeritus
What does L'Hopital's rule say? Did you apply it?

4. Nov 5, 2013

### Dick

I think we need to show that it is a 0/0 form before applying l'Hopital. It's not quite obvious.

5. Nov 5, 2013

### Qube

I get 10x/10x = 0/0.

6. Nov 5, 2013

### Qube

It's 1, because of L'Hopital's rule.

Is the limit of a product the product of the limits of the terms?

7. Nov 5, 2013

### Qube

Alright, I think I got it! And yes it appears that:

"The limit of a product is the product of the limits."

Once again, hats off to the amazingly talented people here at PF. I forgot about the indeterminate forms, and how it is erroneous to conclude anything about a limit when you have an indeterminate form such as 0(infinity) in the case of 10x * cot(10x) with x tending toward 0. I remember my calculus teacher drilling that into us 2 years ago, asking each of us individually whether a limit existed if we plugged numbers and got 0/0. Needless to say, some people insisted it did not exist, when in fact the answer is correctly "I don't know" (it's not possible to conclude anything from 0/0 or any of the other indeterminate forms)!

https://scontent-b-dfw.xx.fbcdn.net/hphotos-prn2/v/1461024_10201007108881643_1917921054_n.jpg?oh=2fac78e5000f1af4ae35dd0de2890a00&oe=527BFFF8

8. Nov 5, 2013

### Dick

No, no, no. The LIMIT of (10x)/sin(10x) is 1. That doesn't mean (10x)/sin(10x)=1. That's wrong. The purpose of the initial limit was just to show you had a 0/0 form so you are allowed to use l'Hopital. Now you have to actually apply l'Hopital to the original function. And I'd suggest you rearrange it a bit before you do that. Otherwise it will get nasty.

9. Nov 5, 2013

### Qube

What about the idea of the product of limits?

Also I think I did what you said.

10. Nov 5, 2013

### Dick

The product of limits is fine to show it's 0/0 to begin with. Here's a sample of what CAN go wrong. Think about ((1+x)*(1+2x)-1)/x as x->0. The limit of (1+x) and (1+2x) are 1. So the limit of the ratio is (1-1)/0 which is 0/0. But you can't substitute 1 for (1+x) and conclude the original limit is the same as ((1+2x)-1)/x=2. That's wrong. The original limit was 3.

Last edited: Nov 6, 2013