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anonymous
 3 years ago
Use L'hospital rule to find the following limit:
Lim x>0+ 6(tan(4x))^x
I did this: Lim x>0+ xln 6(tan(4x))
which is the same as: ln 6(tan(4x)) / (1/x)
Take the derivative: (1/6tan(4x))) * 6(sec^2(4x))(4) / (1/x^2).
But I can't find a way to cancel out my denominator :/ the answer isn't infinity either
anonymous
 3 years ago
Use L'hospital rule to find the following limit: Lim x>0+ 6(tan(4x))^x I did this: Lim x>0+ xln 6(tan(4x)) which is the same as: ln 6(tan(4x)) / (1/x) Take the derivative: (1/6tan(4x))) * 6(sec^2(4x))(4) / (1/x^2). But I can't find a way to cancel out my denominator :/ the answer isn't infinity either

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0leave the six out of it, put it in at the end

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x\ln(\tan(4x)\] \[\frac{\ln(\tan(4x))}{\frac{1}{x}}\]etc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[4x^2\csc(4x)\sec(4x)\] i think is the next step

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i see still a problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe flip the other way

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large= \frac{ 4x }{ \sin 4x }\frac{ x }{ \cos 4x }\] there is no problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your sin will make the denominator 0 even if multiplied :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or am I just not understanding something ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no @sirm3d as it! first limit is 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.00ooh take the limit of each... stupid me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks to both of you
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