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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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Please, don't formulate your questions as if you are giving us homework. Show us what you tried and show us where you got stuck. –  sxd Nov 20 '11 at 1:32
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Immediate required reading (It does a body good.): Homework-question FAQ. –  cardinal Nov 20 '11 at 1:39
    
Can you find upper bounds for each term? –  marty cohen Nov 20 '11 at 2:11

2 Answers 2

up vote 0 down vote accepted

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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tanks! I try to work .... –  FrConnection Nov 21 '11 at 13:51
    
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 $$-\Biggl(k+\frac{1}{k}\Biggr)^n\le x_n\le\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n.$$ $$\lim_{n\to \infty}-\Biggl(k+\frac{1}{k}\Biggr)^n=-\infty$$ $$\lim_{n\to \infty}\Biggl(\frac{k^2+n^2}{kn^2}\Biggr)^n=0$$ $$x_n=\nexists$$ –  FrConnection Nov 21 '11 at 17:55

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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no i don't know Taylor series. –  FrConnection Nov 20 '11 at 14:25
    
@FrConnection: do you know that $\sin x \approx x$ for $x \ll 1$? Similarly $\cos x \approx 1-x^2/2$ for $x \ll 1$? If not, what tools do you have? –  Ross Millikan Nov 20 '11 at 16:23
    
only significant limits and theorems on sequences –  FrConnection Nov 20 '11 at 16:30

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