91Ó°ÊÓ

The sequence is given by: $$a_n = n(1-\cos(\frac{1}{n}))$$ Our goal is to find the limit as \(n\to\infty\) of this sequence, denoted as: $$\lim_{n\to\infty}a_n$$

Rewrite the sequence to be in the form of a fraction: $$a_n = \frac{1-\cos(\frac{1}{n})}{\frac{1}{n}}$$ First, we notice that as \(n\to\infty\), we have \(\cos(\frac{1}{n})\to\cos(0)=1\) and the sequence becomes \(\frac{0}{0}\). This form is an indeterminate one and a candidate for L'Hopital’s rule. L'Hôpital's rule states that if the limit \(\lim_{x\to a}\frac{f(x)}{g(x)}\) is indeterminate (\(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)), then the limit is equal to \(\lim_{x\to a}\frac{f'(x)}{g'(x)}\), provided that the latter limit exists. Now, differentiate the numerator and the denominator with respect to \(n\): $$ f(n) = 1-\cos(\frac{1}{n}) \\ g(n) = \frac{1}{n} $$ $$ f'(n) = \frac{1}{n^2}\sin(\frac{1}{n}) \\ g'(n) = -\frac{1}{n^2} $$

Now applying L'Hopital's rule to find the limit as \(n\to\infty\), we get: $$\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{(\frac{1}{n^2}\sin(\frac{1}{n}))}{(-\frac{1}{n^2})} $$ Simplify the above expression: $$\lim_{n\to\infty}\frac{1}{n^2}\sin(\frac{1}{n})\cdot\frac{-n^2}{1} = \lim_{n\to\infty}(-\sin(\frac{1}{n}))$$

As \(n\to\infty\), the sequence converges to the limit: $$\lim_{n\to\infty}(-\sin(\frac{1}{n})) = -\sin(0) = 0$$ Thus, the limit of the given sequence is 0.