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sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361572558530:dw

sat_chen
 one year 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

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0so i use the rule and get dw:1361572627877:dw

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

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0the derivative of tan is sec^2 x lol

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

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

calmat01
 one year 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.

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

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0you can change it into tan^2x though, not tan^2x

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0ok im gonna try walters method

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0But that only makes it worse. Nevermind. Hang on.

sat_chen
 one year 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

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0Move the 1 over, and then multiply both sides by a 1.

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

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

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0so the solution is 1/0 but thats not gonna work either

experimentX
 one year 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

calmat01
 one year 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)

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0no that is the question experimentX

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0oops, it's approaching zero, so the answer is 1/2

experimentX
 one year 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.

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0Ah hell, then it's no longer of indeterminate form.

experimentX
 one year 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.

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0i think i got it now ty a lot

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0I need to go refresh my memory. Good luck to you.
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