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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
 one year ago
 one year 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
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.0
leave the six out of it, put it in at the end
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
\[x\ln(\tan(4x)\] \[\frac{\ln(\tan(4x))}{\frac{1}{x}}\]etc
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
\[4x^2\csc(4x)\sec(4x)\] i think is the next step
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
oh i see still a problem
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
maybe flip the other way
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
\[\large= \frac{ 4x }{ \sin 4x }\frac{ x }{ \cos 4x }\] there is no problem
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
your sin will make the denominator 0 even if multiplied :/
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
or am I just not understanding something ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
no @sirm3d as it! first limit is 1
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
0ooh take the limit of each... stupid me.
 one year ago

MarcLeclairBest ResponseYou've already chosen the best response.0
Thanks to both of you
 one year ago
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