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Prove the following two limits using the definition of limit of a sequence.

$$\lim_{n\rightarrow \infty} \frac{n}{n^2+1} = 0$$

$$\lim_{n\rightarrow \infty} \frac{n^2 -1}{2n^2 +3} = \frac{1}{2}$$

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  • $\begingroup$ Let epsilon > 0 be arbitrary. I must show that there exists an N that is an element of the Natural numbers such that n >= N implies abs( (n / n^2 + 1) - 0) < epsilon. This is for the first proof of course $\endgroup$ – user123652 Feb 18 '14 at 4:10
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For the first problem, note that

$$\left|\frac{n}{n^{2}+1} - 0\right|= \left|\frac{n}{n^{2}+1}\right|=\frac{n}{n^{2}+1}$$

— the last piece true since $n\in \mathbb{N}$. Therefore, we need $\frac{n}{n^2+1} < \epsilon$ in order to show that the sequence converges to $0$. Note also that

$$\frac{n}{n^2+1}<\frac{n}{n^2}$$

for all $n\in \mathbb{N}$. Therefore, if we make the latter term less than $\epsilon$...

Do you see the general idea of how to proceed, how we choose $N$? Let me know if you have questions.

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