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

Constantine

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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:

limx0(1+sin(4x))cot(x)\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\, \Big(1\, +\, \sin(4x)\Big)^{\cot(x)}

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

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

limx0ln(y)=limx0ln(1+sin(4x))tan(x)=limx0(4cos(4x)1+sin(4x))sec2(x)=4\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

limx0(1+sin(4x))cot(x)=e4\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
 

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

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