anonymous
  • anonymous
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
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
leave the six out of it, put it in at the end
anonymous
  • anonymous
\[x\ln(\tan(4x)\] \[\frac{\ln(\tan(4x))}{\frac{1}{x}}\]etc
anonymous
  • anonymous
\[-4x^2\csc(4x)\sec(4x)\] i think is the next step

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anonymous
  • anonymous
oh i see still a problem
anonymous
  • anonymous
maybe flip the other way
sirm3d
  • sirm3d
\[\large= \frac{ 4x }{ \sin 4x }\frac{ -x }{ \cos 4x }\] there is no problem
anonymous
  • anonymous
your sin will make the denominator 0 even if multiplied :/
anonymous
  • anonymous
or am I just not understanding something ?
anonymous
  • anonymous
no @sirm3d as it! first limit is 1
anonymous
  • anonymous
*has it
anonymous
  • anonymous
0ooh take the limit of each... stupid me.
anonymous
  • anonymous
Thanks to both of you

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