A huge limit.

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

lim⁡x→0ln⁡sin⁡2nxsin⁡xn−ln⁡2ln⁡(e+sin⁡2x)(e+sin⁡22x)...(e+sin⁡2nx)n−1 \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

#Calculus #Limits

Note by Haroun Meghaichi
6 years, 11 months ago

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1 vote

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Comments

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

lim⁡x→0ln⁡(cos⁡x×cos⁡2x×cos⁡4x×...×cos⁡2n−1x)ln⁡(1+sin⁡2xe)(1+sin⁡22xe)(1+sin⁡23xe)...(1+sin⁡2nxe)\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⁡x→0ln⁡(1+cos⁡x−1)cos⁡x−1×cos⁡x−1x2+ln⁡(1+cos⁡2x−1)cos⁡2x−1×cos⁡2x−1(2x)2×22+...+ln⁡(1+cos⁡2n−1x−1)cos⁡2n−1x−1×cos⁡2n−1x−1(2n−1x)2×22n−2ln⁡(1+sin⁡2xe)sin⁡2xe×sin⁡2xex2+ln⁡(1+sin⁡22xe)sin⁡22xe×sin⁡22xe(2x)2×22+...+ln⁡(1+sin⁡2nxe)sin⁡2nxe×sin⁡2nxe(nx)2×n2=\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}}

=−12(20+22+24+...22n−2)1e(12+22+32+...+n2)=\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})}

=(1−22n)en(n+1)(2n+1)=\dfrac{(1-2^{2n})e}{n(n+1)(2n+1)}

Karthik Kannan - 6 years, 11 months ago

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As the lowest power of xx in expansion of denominator is 22, we can put ln⁡(1+cos⁡2rx−1)≈cos⁡2rx−1≈−122rx2\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 :)

jatin yadav - 6 years, 11 months ago

Using series expansion, I get the answer as (1−22n)en(n+1)(2n+1)\dfrac{(1 - 2 ^{2n}) e}{n(n+1)(2n+1)} . Is it correct?

jatin yadav - 6 years, 11 months ago

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I believe that you should get ee instead of e2e^2.

Haroun Meghaichi - 6 years, 11 months ago

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Yes, I meant ee, edited.

jatin yadav - 6 years, 11 months ago
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