find this limit using the l hopital rule

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find this limit using the l hopital rule

Mathematics
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|dw:1361572558530:dw|
i know if you just plug in the value you get 0/0 so you use the rule
so i use the rule and get |dw:1361572627877:dw|

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how do i get from here as you cant use the trigonometric substitution either
|dw:1361572546361:dw|
the derivative of tan is sec^2 x lol
so thats wrong
|dw:1361572818037:dw| then differentiate from here
the limit goes to -infinity
Change your 1 -sec^2x into -tan^2x and see if that helps you simplify your ratio.
thats not how it works youcan never change 1 - sec^2 into tan^2x
you can change it into -tan^2x though, not tan^2x
ok im gonna try walters method
But that only makes it worse. Nevermind. Hang on.
you dont understand tan^2 x + 1 = sec^2x try to change it to what you just did it never works
Move the 1 over, and then multiply both sides by a -1.
i thought about your method walter but its x not 1 - tanx
Without using L'hopital |dw:1361573289043:dw|
so the solution is 1/0 but thats not gonna work either
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
If you use L'hopital's rule twice, you end up with the limit as x approaches negative infinity of cos^3x/(-2)
no that is the question experimentX
oops, it's approaching zero, so the answer is -1/2
|dw:1361573583223:dw|
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.
*once
Ah hell, then it's no longer of indeterminate form.
|dw:1361573754561:dw|
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.
i think i got it now ty a lot
I need to go refresh my memory. Good luck to you.

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