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

sat_chenBest ResponseYou've already chosen the best response.0
i know if you just plug in the value you get 0/0 so you use the rule
 one year ago

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

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

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

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

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

calmat01Best ResponseYou've already chosen the best response.0
Change your 1 sec^2x into tan^2x and see if that helps you simplify your ratio.
 one year ago

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

calmat01Best ResponseYou've already chosen the best response.0
you can change it into tan^2x though, not tan^2x
 one year ago

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

calmat01Best ResponseYou've already chosen the best response.0
But that only makes it worse. Nevermind. Hang on.
 one year ago

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

calmat01Best ResponseYou've already chosen the best response.0
Move the 1 over, and then multiply both sides by a 1.
 one year ago

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

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

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

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

calmat01Best ResponseYou've already chosen the best response.0
If you use L'hopital's rule twice, you end up with the limit as x approaches negative infinity of cos^3x/(2)
 one year ago

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

calmat01Best ResponseYou've already chosen the best response.0
oops, it's approaching zero, so the answer is 1/2
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
Ah hell, then it's no longer of indeterminate form.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

calmat01Best ResponseYou've already chosen the best response.0
I need to go refresh my memory. Good luck to you.
 one year ago
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