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Find the limit $\displaystyle{\lim_{x\to0}(1+2x)^{3\csc(2x)}}$

I did the following $(1+2x)^{3\csc(2x)}=e^{\ln(1+2x)3\csc(2x)}$, took $\ln$ on the lim getting

$3\frac{\ln(1+2x)}{\frac{1}{\csc(2x)}}$

I did hospital rule

and got

$\lim3\cdot2\frac{\frac{1}{2x}}{\frac{-1}{\csc(2x)cot(2x) 2}}$

then I find myself stuck on what to do.

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  • $\begingroup$ Similar questions were asked and answered a lot. The Maple command $$ Student[Calculus1]:-LimitTutor((1+2*x)^{3*csc(2*x)}, x = 0) $$ finds it step by step with explanations. See here and here. $\endgroup$ – user64494 Oct 30 '13 at 19:11
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$$ \frac{\ln(1+2x)}{\frac1{\csc 2x}}=\frac{\ln(1+2x)}{\sin 2x}. $$ Being a $0/0$ of differentiable functions, L'Hopital applies, so $$ \lim_{x\to0}\frac{\ln(1+2x)}{\frac1{\csc 2x}}=\lim_{x\to0}\frac{\ln(1+2x)}{\sin 2x}=\lim_{x\to0}\frac{\frac2{1+2x}}{2\cos2x}=\lim_{x\to0}\frac1{(1+2x)\cos2x}=1 $$

Or, one can use Taylor: $$ \frac{\ln(1+2x)}{\sin 2x}=\frac{2x+O(x^2)}{2x+O(x^3)}=\frac{1+O(x)}{1+O(x^2)}\to1. $$

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  • $\begingroup$ Yes I forgot 1/csc(2x) is sin(2x) $\endgroup$ – Fernando Martinez Oct 30 '13 at 18:55
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Hint $\lim_{x\to 0}(1+x)^{1\over x}$?

$\lim_{x\to 0}(1+2x)^{3\csc(2x)}=\lim _{x\to 0}[(1+2x)^{1\over 2x}]^{3\times\lim_{x\to 0}{2x\over \sin 2x}}=e^3$

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  • $\begingroup$ that would be e^(1/x) $\endgroup$ – Fernando Martinez Oct 30 '13 at 18:33
  • $\begingroup$ No just $e$...if you want I will add more $\endgroup$ – Marso Oct 30 '13 at 18:34
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$$\lim_{x\to0}(1+2x)^{3\csc(2x)}=\lim_{x\to0}(1+2x)^{\frac{1}{2x}\frac{2x}{\sin(2x)}3}=e^3$$

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