sat_chen 2 years ago find this limit using the l hopital rule

1. sat_chen

|dw:1361572558530:dw|

2. sat_chen

i know if you just plug in the value you get 0/0 so you use the rule

3. sat_chen

so i use the rule and get |dw:1361572627877:dw|

4. sat_chen

how do i get from here as you cant use the trigonometric substitution either

5. walters

|dw:1361572546361:dw|

6. sat_chen

the derivative of tan is sec^2 x lol

7. sat_chen

so thats wrong

8. walters

|dw:1361572818037:dw| then differentiate from here

9. experimentX

the limit goes to -infinity

10. calmat01

Change your 1 -sec^2x into -tan^2x and see if that helps you simplify your ratio.

11. sat_chen

thats not how it works youcan never change 1 - sec^2 into tan^2x

12. calmat01

you can change it into -tan^2x though, not tan^2x

13. sat_chen

ok im gonna try walters method

14. calmat01

But that only makes it worse. Nevermind. Hang on.

15. sat_chen

you dont understand tan^2 x + 1 = sec^2x try to change it to what you just did it never works

16. calmat01

Move the 1 over, and then multiply both sides by a -1.

17. sat_chen

i thought about your method walter but its x not 1 - tanx

18. experimentX

Without using L'hopital |dw:1361573289043:dw|

19. sat_chen

so the solution is 1/0 but thats not gonna work either

20. experimentX

or I could have misunderstood your question http://www.wolframalpha.com/input/?i=lim+x-%3E0+sin%28x%29%2F%28x-tan%28x%29%29

21. calmat01

If you use L'hopital's rule twice, you end up with the limit as x approaches negative infinity of cos^3x/(-2)

22. sat_chen

no that is the question experimentX

23. calmat01

oops, it's approaching zero, so the answer is -1/2

24. walters

|dw:1361573583223:dw|

25. experimentX

You can't use L'hopital rule twice, one you use L'hopital rule, ... the top is cos(x) ... which is 1 and the bottom part is still 0.

26. experimentX

*once

27. calmat01

Ah hell, then it's no longer of indeterminate form.

28. walters

|dw:1361573754561:dw|

29. experimentX

the best way see through limits is to expand using Taylor series. Try expanding it ... the answer is pretty obvious. Anyway gotta sleep ... best of luck.

30. sat_chen

i think i got it now ty a lot

31. calmat01

I need to go refresh my memory. Good luck to you.