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anonymous
 3 years ago
find this limit using the l hopital rule
anonymous
 3 years ago
find this limit using the l hopital rule

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361572558530:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know if you just plug in the value you get 0/0 so you use the rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i use the rule and get dw:1361572627877:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do i get from here as you cant use the trigonometric substitution either

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the derivative of tan is sec^2 x lol

walters
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361572818037:dw then differentiate from here

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the limit goes to infinity

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Change your 1 sec^2x into tan^2x and see if that helps you simplify your ratio.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats not how it works youcan never change 1  sec^2 into tan^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can change it into tan^2x though, not tan^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok im gonna try walters method

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But that only makes it worse. Nevermind. Hang on.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you dont understand tan^2 x + 1 = sec^2x try to change it to what you just did it never works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Move the 1 over, and then multiply both sides by a 1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought about your method walter but its x not 1  tanx

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Without using L'hopital dw:1361573289043:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the solution is 1/0 but thats not gonna work either

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1or I could have misunderstood your question http://www.wolframalpha.com/input/?i=lim+x%3E0+sin%28x%29%2F%28xtan%28x%29%29

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you use L'hopital's rule twice, you end up with the limit as x approaches negative infinity of cos^3x/(2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no that is the question experimentX

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oops, it's approaching zero, so the answer is 1/2

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1You 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah hell, then it's no longer of indeterminate form.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the 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
 3 years ago
Best ResponseYou've already chosen the best response.0i think i got it now ty a lot

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I need to go refresh my memory. Good luck to you.
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