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Good day, I'm not sure that this limit exists. All my attempts to prove it were in vain ...

Let $k>1$. If exist, calculate the limit of the sequence $(x_n)$ defined by, $x_n := \Biggl(k \sin \left(\frac{1}{n^2}\right) + \frac{1}{k}\cos n \Biggr)^n.$

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    Can you find upper bounds for each term?2011-11-20

2 Answers 2

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HINT: For all $n\in\mathbb{Z}^+$ you have $\sin\frac1{n^2}\le\frac1{n^2}\text{ and }\frac1k\cos n\le\frac1k,$ so $x_n\le\left(\frac{k}{n^2}+\frac1k\right)^n=\left(\frac{k^2+1}{kn^2}\right)^n=\left(\frac{k+\frac1k}{n^2}\right)^n.$ (Be a little careful, though: you still have to worry about a lower bound for the $x_n$.)

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    is correct so? : 0< \sin x < x <\tan x if 0 < x < \pi/2 ; \sin \frac{1}{n^2}<\frac{1}{n^2}; $-\frac{1}{k}\le\cos n \le\frac{1}{k} $ then $x_n \le \Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n;$ $-{k}\le\ ksin (\frac{1}{n^2})\le{k},$ so $-\Biggl(k+\frac{1}{k}\Biggr)^n\le x_n\le\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n.$ $\lim_{n\to \infty}-\Biggl(k+\frac{1}{k}\Biggr)^n=-\infty$ $\lim_{n\to \infty}\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n=0$ $x_n=\nexists$2011-11-21
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Hint: you can replace the trig functions with the first part of their Taylor series. How far do you have to go?

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    only significant limits and theorems on sequences2011-11-20