Indeterminate Powers: lim [x -> 0] (1+sin(4x))^(cot(x))

[Constantine]

Hello, I have a question regarding using l'Hospital's Rule to solve problems with indeterminate powers. The specific problem which comes from my textbook is given below:

\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\, \Big(1\, +\, \sin(4x)\Big)^{\cot(x)}\)

\(\displaystyle \ln(y)\, =\, \ln\Big[(1\, +\, \sin(4x))^{\cot(x)}\Big]\)

\(\displaystyle \ln(y)\, =\, \cot(x)\, \ln(1\, +\, \sin (4x))\)

\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\, \ln(y)\, =\, \lim_{x\, \rightarrow\, 0}\, \dfrac{\ln(1\, +\, \sin (4x))}{\tan(x)}\, =\, \lim_{x\, \rightarrow\, 0}\, \dfrac{\left(\dfrac{4\, \cos(4x)}{1\, +\, \sin(4x)}\right)}{\sec^2(x)}\,=\, 4\)

\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\, \Big(1\, +\, \sin(4x)\Big)^{\cot(x)}\, =\, e^4\)

I understand how to use the Rule but what I don't get is how cotx function from step #3 became tanx function found in the denominator in the following step. I have tried solving the problem numerous times on my own but I don't understand how tanx got there. If somebody could please explain that to me, I would be very grateful.

Thank you,

Constantine

 

[lookagain]

\(\displaystyle cot(x) \ = \ \dfrac{1}{tan(x)}\)

So, \(\displaystyle \ cot(x)*ln[1 + sin(4x)] \ = \ \dfrac{1}{tan(x)}*\dfrac{ln[1 + sin(4x)]}{1} \ = \ \dfrac{ln[1 + sin(4x)]}{tan(x)}\)

 

[Constantine]

That's all that there was to it? I feel humbled. Thank you for your help.