sat_chen
find this limit using the l hopital rule
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sat_chen
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|dw:1361572558530:dw|
sat_chen
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i know if you just plug in the value you get 0/0 so you use the rule
sat_chen
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so i use the rule and get |dw:1361572627877:dw|
sat_chen
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how do i get from here as you cant use the trigonometric substitution either
walters
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|dw:1361572546361:dw|
sat_chen
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the derivative of tan is sec^2 x lol
sat_chen
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so thats wrong
walters
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|dw:1361572818037:dw|
then differentiate from here
experimentX
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the limit goes to -infinity
calmat01
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Change your 1 -sec^2x into -tan^2x and see if that helps you simplify your ratio.
sat_chen
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thats not how it works youcan never change 1 - sec^2 into tan^2x
calmat01
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you can change it into -tan^2x though, not tan^2x
sat_chen
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ok im gonna try walters method
calmat01
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But that only makes it worse. Nevermind. Hang on.
sat_chen
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you dont understand tan^2 x + 1 = sec^2x
try to change it to what you just did it never works
calmat01
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Move the 1 over, and then multiply both sides by a -1.
sat_chen
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i thought about your method walter but its x not 1 - tanx
experimentX
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Without using L'hopital
|dw:1361573289043:dw|
sat_chen
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so the solution is 1/0 but thats not gonna work either
calmat01
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If you use L'hopital's rule twice, you end up with the limit as x approaches negative infinity of cos^3x/(-2)
sat_chen
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no that is the question experimentX
calmat01
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oops, it's approaching zero, so the answer is -1/2
walters
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|dw:1361573583223:dw|
experimentX
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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
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*once
calmat01
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Ah hell, then it's no longer of indeterminate form.
walters
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|dw:1361573754561:dw|
experimentX
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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.
sat_chen
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i think i got it now ty a lot
calmat01
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I need to go refresh my memory. Good luck to you.