I'm asked to say what the following limit is, and then prove it using the definition of convergence.
$\lim_{n\rightarrow\infty}$$\dfrac{3n^2+1}{4n^2+n+2}$.
Is it valid to say that the sequence behaves like $\dfrac{3n^2}{4n^2}$ for large n?
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Not if you want to do this formally. – Akiva Weinberger Apr 01 '15 at 01:55
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Your statement is valid, but it doesn't use the definition of convergence (i.e., the $\epsilon-N$ definition of convergence). To prove that the sequence converges to $3/4$, consider
$$\left|\frac{3n^2 + 1}{4n^2 + n + 2} - \frac{3}{4}\right| = \left|\frac{-3n - 2}{4(4n^2 +n + 2)}\right| = \frac{3n + 2}{4(4n^2 + n + 2)} < \frac{3n + 2n}{16n^2} = \frac{5}{16n} \tag{*}$$
Given $\epsilon > 0$, setting $N > 5/(16\epsilon)$ will make the rightmost side (and hence the leftmost side) of $(*)$ less than $\epsilon$ for all $n\ge N$. Therefore,
$$\lim_{n\to \infty} \frac{3n^2 + 1}{4n^2 + n + 2} = \frac{3}{4}.$$
kobe
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Just to expand a bit on the above solution, just to the left of the first (*) we need $\frac{5}{16n}< ϵ $ which implies that we need $\frac{5}{16ϵ} < n$ in case it wasn't already plainly obvious. Then setting N > $\frac{5}{16ϵ}$ guarantees the result for all n≥N.
SWilliams
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