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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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    Try factoring $n^2$ out of the numerator and denominator, then work with the remaining pieces. – 2017-02-28
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    Guys I know that, but I'm trying to prove it by definition. As in finding an n that depends on epsilon. – 2017-02-28
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    My comment remains the same...instead of trying to do it directly, work with things that go to $0$ instead of $\infty$... – 2017-02-28
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    Ok I tried it, but I'm still having difficulty getting it simplified enough. I just ended up with more fractions. – 2017-02-28

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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.

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    Thank you! Saved my assignment :D – 2017-02-28
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    Do you understand the proof? Could you now do this for a different function? – 2017-03-05
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Given any $\epsilon \in \mathbb{R}^+$, assuming $n>10$ we want $$ \left|\frac{n^2+n+1}{2 n^2-4 n+1}-\frac{1}{2}\right|=\left|\frac{6 n+1}{4 n^2-8 n+2}\right|\le\left|\frac{7 n}{4 n^2-6 n}\right|=7\left|\frac{1}{4 n-6 }\right|\le\frac{7}{n}\lt\epsilon $$ Means $$ n>\max\left\{10,\frac{7}{\epsilon}\right\} \implies \left|\frac{n^2+n+1}{2 n^2-4 n+1}-\frac{1}{2}\right|\le \epsilon $$