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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

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

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

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

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

Log in to reply

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 :)

Log in to reply

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

Log in to reply

I believe that you should get \(e\) instead of \(e^2\).

Log in to reply

Yes, I meant \(e\), edited.

Log in to reply