Use I'Hopital's rule to find the limit of the sequence
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That is overkilling it. $\displaystyle n\sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}}$.
thanks for your prompt answer, but how did pie get in?
Hello, Originally Posted by al1850 thanks for your prompt answer, but how did pie get in? Where do you see any pie ? Or any $\displaystyle \pi$ ?
Hi Maybe in the ThePerfectHacker's signature ? $\displaystyle \boxed{ -\sin \frac{\pi}{7} + \sin \frac{2\pi}{7} + \sin \frac{3\pi}{7} + \sin \frac{4\pi}{7}+\sin \frac{5\pi}{7} - \sin \frac{6\pi}{7} = \sqrt{7} } $
Haha ! That's funny
oh man, that's your signature? I thought that was the answer to the I'Hopital's rule question. lol
So is (sin 1/n)/(1/n) the answer?
$\displaystyle \lim_{n\to\infty}\frac{\sin\frac{1}{n}}{\frac{1}{n }}$ Substitute $\displaystyle a=\frac{1}{n}$.
And remember the definition of the derivative : $\displaystyle \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$ Here, $\displaystyle f(x)=\sin(x)$ Take $\displaystyle x_0=0$ Or else, apply l'Hôpital's rule
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