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# A huge limit.

Let $$n\geq1$$ be an integer, calculate the following limit

$\lim_{x\to0}\frac{\ln\sqrt[n]{\frac{\sin2^nx}{\sin x}}-\ln2}{\ln\sqrt[n]{(e+\sin^2x)(e+\sin^22x)...(e+\sin^2nx)}-1}$

Source : I've encountered this limit a couple years ago, I think that it is from some journal but I'm not sure

Note by Haroun Meghaichi
3 years ago

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After a little manipulation the given limit can be written as:

$$\lim\limits_{x\to 0} \displaystyle\frac{\ln (\cos x\times\cos 2x\times\cos 4x\times...\times\cos 2^{n-1}x)}{\ln \left( 1+\dfrac{\sin^{2} x}{e}\right)\left( 1+\dfrac{\sin^{2} 2x}{e}\right)\left( 1+\dfrac{\sin^{2} 3x}{e}\right)...\left( 1+\dfrac{\sin^{2} nx}{e}\right)}$$

$$=\lim\limits_{x\to 0} \displaystyle\frac{\dfrac{\ln (1+\cos x-1)}{\cos x-1}\times\dfrac{\cos x-1}{x^{2}}+ \dfrac{\ln (1+\cos 2x-1)}{\cos 2x-1}\times\dfrac{\cos 2x-1}{(2x)^{2}}\times 2^{2}+...+\dfrac{\ln (1+\cos 2^{n-1}x-1)}{\cos 2^{n-1}x-1}\times\dfrac{\cos 2^{n-1}x-1}{(2^{n-1}x)^{2}}\times 2^{2n-2}}{\dfrac{\ln \left( 1+\dfrac{\sin^{2} x}{e}\right)}{\dfrac{\sin^{2} x}{e}}\times\dfrac{\dfrac{\sin^{2} x}{e}}{x^{2}}+\dfrac{\ln \left( 1+\dfrac{\sin^{2} 2x}{e}\right)}{\dfrac{\sin^{2} 2x}{e}}\times\dfrac{\dfrac{\sin^{2} 2x}{e}}{(2x)^{2}}\times 2^{2}+...+\dfrac{\ln \left( 1+\dfrac{\sin^{2} nx}{e}\right)}{\dfrac{\sin^{2} nx}{e}}\times\dfrac{\dfrac{\sin^{2} nx}{e}}{(nx)^{2}}\times n^{2}}$$

$$=\displaystyle\frac{\dfrac{-1}{2}(2^{0}+2^{2}+2^{4}+...2^{2n-2})}{\dfrac{1}{e}(1^{2}+2^{2}+3^{2}+...+n^{2})}$$

$$=\dfrac{(1-2^{2n})e}{n(n+1)(2n+1)}$$ · 3 years ago

As the lowest power of $$x$$ in expansion of denominator is $$2$$, we can put $$\ln(1 + \cos 2^rx - 1) \approx \cos 2^rx - 1 \approx - \dfrac{1}{2} 2^r x^2$$. Hence, we avoid lengthy seeming expression :) · 3 years ago

Using series expansion, I get the answer as $$\dfrac{(1 - 2 ^{2n}) e}{n(n+1)(2n+1)}$$ . Is it correct? · 3 years ago

I believe that you should get $$e$$ instead of $$e^2$$. · 3 years ago

Yes, I meant $$e$$, edited. · 3 years ago