## anonymous 5 years ago How does sin(t)/(1-cos(t)) simplify to cos(t)/sin(t)?

1. anonymous

are you sure that's the exact question as written?

2. anonymous

It doesn't. Pick t=5 and you'll see it doesn't work.

3. anonymous

For example...

4. anonymous

Well it's that I'm looking at a solution my teacher wrote. He was showing how a cycloid had infinitely sharp cusps by finding the limit of the derivative of the parametric formulas given. So he was trying first to find the limit of sin(t)/(1-cos(t)) and he jumped to trying to find the limit of cos(t)/sin(t) because apparently, they're equal.

5. anonymous

Look at #5. http://people.sfcollege.edu/bruce.teague/pdf/2312%20W11/2312_MG5_solutions_W11.pdf This link should work.

6. anonymous

Did everybody leave?

7. anonymous

Awww c'mon!

8. anonymous

Now what am I gonna do?

9. anonymous

Calm down...I'll have a look.

10. anonymous

Haha ok

11. anonymous

Oh, he's used L'Hopital's rule...do you know about that?

12. anonymous

Maybe. I might have forgotten.

13. anonymous

14. anonymous

When you have indeterminate forms for the limit (numerator and denominator each go to something like 0/0 or infinity/infinity), the limit of the original ratio is equal to the limit ratio of the derivatives of the numerator and denominator.

15. anonymous

Take the derivative of the numerator of sin(t) and you get cos(t), and the derivative of 1-cost(t) is sin(t).

16. anonymous

Oh! It's all coming back now.

17. anonymous

OK. That makes sense. Thank you soooooo much.

18. anonymous

Fan me then! :) I want to reach superstar!

19. anonymous

How does one fan someone else?

20. anonymous

$\lim_{t \rightarrow 0} \frac{sin(t)}{1-\cos(t)}= \frac{0}{1-1}=\frac{0}{0}$ lokisan is right, this is an indeterminate and using L'hopital would give you

21. anonymous

there should be a link-type thing next to my name..."Become a fan"

22. anonymous

I'm not seeing it. I'm looking.

23. anonymous

It's in the thread window...little 'thumbs up' icon. There's one next to your own name too.

24. anonymous

There. I got it.

25. anonymous

Awesome...cheers!