Good day, I was wondering how I could prove that
$$\lim_{n\to \infty}\frac{n^2+n+1}{2n^2-4n+1}=1/2$$
By definition.
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$\begingroup$You need to show that for every $\epsilon>0$ we can find $N_\epsilon\in\mathbb{N}$ such that when $n>N_\epsilon$, we have
$$\left|\frac{n^2+n+1}{2n^2-4n+1}-\frac{1}{2}\right|<\epsilon$$
Simplifying a bit, we want
$$\left|\frac{6n+1}{2n^2-4n+1}\right|=\frac{6n+1}{|2n^2-4n+1|}<2\epsilon$$
Since $|2n^2-4n+1|>n^2$ when $n>5$, we have
$$\frac{6n+1}{|2n^2-4n+1|}<\frac{6n+1}{n^2}=\frac{6}{n}+\frac{1}{n^2}<\frac{6}{n}+\frac{1}{n}=\frac{7}{n}$$
If $\frac{7}{n}$ is to be smaller than $2\epsilon$, we need $n>\frac{7}{2\epsilon}$. So taking $N_\epsilon=\max\left\{5, \frac{7}{2\epsilon}\right\}$ will suffice.
$\endgroup$ 2 $\begingroup$Given any $\epsilon \in \mathbb{R}^+$, assuming $n\ge3$ we have$$ \left|\frac{n^2+n+1}{2 n^2-4 n+1}-\frac{1}{2}\right|=\left|\frac{6 n+1}{4 n^2-8 n+2}\right|\le\left|\frac{7 n}{4 n^2-6 n}\right|=7\left|\frac{1}{4 n-6 }\right|\le\frac{7}{n}\lt\epsilon $$Means$$ n>\max\left\{3,\frac{7}{\epsilon}\right\} \implies \left|\frac{n^2+n+1}{2 n^2-4 n+1}-\frac{1}{2}\right|< \epsilon $$
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