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We've got a question that shows us how to find the limit of this sequence as the nth term approaches infinity. I'm unsure of if we should use L'Hopitals rule, or if not what we should use instead. I can see, we may be able to use L'hopital as the formula will be infinity/infinity.

We have that

$$a_n=\frac{n\cos(n\pi+\pi/3)+n(-1)^n}{n^2+1}$$

Then, how evaluate $\lim_{n\to\infty}a_n$?

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Hint: $|\cos(x)| \le 1$................

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Hint: use the fact that $a_n=\frac{n}{n^2+1}\cdot b_n=\frac{b_n}{n}\cdot \frac{1}{1+\frac{1}{n^2}}$ where $b_n$ is a bounded sequence because $$|b_n|=|\cos(n\pi+\pi/3)+(-1)^n|\leq |\cos(n\pi+\pi/3)|+|(-1)^n|\leq 2.$$ Note that $0\leftarrow-\frac{2}{n}\leq\frac{b_n}{n}\leq \frac{2}{n}\rightarrow 0$ as $n\to \infty$.

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