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

Question

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 · Answer

leave the six out of it, put it in at the end

anonymous · Answer

$$x\ln(\tan(4x)$$
$$\frac{\ln(\tan(4x))}{\frac{1}{x}}$$etc

anonymous · Answer

$$-4x^2\csc(4x)\sec(4x)$$ i think is the next step

anonymous · Answer

oh i see still a problem

anonymous · Answer

maybe flip the other way

sirm3d · Answer

$$\large= \frac{ 4x }{ \sin 4x }\frac{ -x }{ \cos 4x }$$ there is no problem

anonymous · Answer

your sin will make the denominator 0 even if multiplied :/

anonymous · Answer

or am I just not understanding something ?

anonymous · Answer

no @sirm3d  as it! first limit is 1

anonymous · Answer

*has it

anonymous · Answer

0ooh take the limit of each... stupid me.

anonymous · Answer

Thanks to both of you