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Please, help me in solving of $\lim\limits_{m\to\infty}\left(\cos\frac xm\right)^{m}$.

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l'Hôpital's rule –  Elements in Space Jan 23 '13 at 8:29
    
Just notice this $\cos(\frac{x}{m})\sim 1$ as $ m\to \infty $. –  Mhenni Benghorbal Jan 26 '13 at 5:02
    
@MhenniBenghorbal Just notice this... Exactly the mistake to avoid. Note that $1\sim1$, $1+1/m\sim1$ and $1+1/\sqrt{m}\sim1$ while $1^m\to1$, $(1+1/m)^m\to\mathrm e$ and $(1+1/\sqrt{m})^m\to\infty$. –  Did Jan 26 '13 at 5:09
    
@MhenniBenghorbal I thought about it, but $1^\infty$ is still indeterminate in that case. –  Igor Gorbunov Jan 27 '13 at 12:35

3 Answers 3

up vote 3 down vote accepted

The approximation $\cos(x)\approx 1-\frac{x^2}{2}$ (which has several geometric proofs) and the binomial approximation $(1+t)^n\approx 1+nt$ are enough to give you $\displaystyle\left(\cos\frac{x}{m}\right)^m\approx\left(1-\frac{x^2}{2m^2}\right)^m\approx 1-\frac{x^2}{2m}$, and of course taking the limit as $m\to\infty$ in the latter expression is trivial.

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$1\approx 2$ but they don't have equal limits. –  Michael Albanese Jan 23 '13 at 8:44
    
Thank you! This is fine. –  Igor Gorbunov Jan 23 '13 at 9:02
    
Nice and fine answer +1 –  B. S. Jan 23 '13 at 9:05
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@MichaelAlbanese That depends entirely on what your usage of $\approx$ is. I'm (informally and implicitly) using a version in which $a\approx b+c$ stands for $a=b+c+o(c)$; it's easy to formalize these manipulations and for a proof one would obviously want to be a lot more careful than I was here, but for a problem this simple I felt the extra notation just gets in the way. –  Steven Stadnicki Jan 23 '13 at 16:36
    
Of course, I should have made myself a bit clearer. I wasn't doubting your method, I just thought the use of the symbol $\approx$ doesn't make it clear that the approximations need to be fairly good ones (in the sense that you have made clear in your comment). –  Michael Albanese Jan 23 '13 at 20:17

Another way $$\lim\limits_{m\to\infty}\left(1+\left(\cos\frac xm-1\right)\right)^{m}=\lim\limits_{m\to\infty}e^{\frac{\left(\displaystyle\cos\frac xm-1\right)}{\displaystyle\left(\frac{x}{m}\right)^2}\times \displaystyle\frac{x^2}{m}}=\lim\limits_{m\to\infty}e^{\displaystyle-\frac{1}{2}\times\frac{x^2}{m}}=e^0=1$$

Done.

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The answer is 1. Try taylor expanding cos and using $\lim\limits_{x\to0} \frac{\log({1+x})}{x}=1$.

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Thank you! But is there a possibility to solve it without using l'Hôpital's rule or Taylor expanding, i.e. using trigonometric identities? –  Igor Gorbunov Jan 23 '13 at 8:34
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Well, instead of taylor expanding you can use $cos(2x)=1-2sin^2(x)$ and $ lim_{x\to 0} \frac{sin(x)}{x}=1$ –  Ishan Banerjee Jan 23 '13 at 8:40

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