find this limit using the l hopital rule

- anonymous

find this limit using the l hopital rule

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- anonymous

|dw:1361572558530:dw|

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- walters

|dw:1361572546361:dw|

- anonymous

the derivative of tan is sec^2 x lol

- anonymous

so thats wrong

- walters

|dw:1361572818037:dw|
then differentiate from here

- experimentX

the limit goes to -infinity

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok im gonna try walters method

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- experimentX

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

- anonymous

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

- 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

- anonymous

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

- anonymous

no that is the question experimentX

- anonymous

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

- walters

|dw:1361573583223:dw|

- 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.

- experimentX

*once

- anonymous

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

- walters

|dw:1361573754561:dw|

- 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.

- anonymous

i think i got it now ty a lot

- anonymous

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

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