Prove that $\lim_{n\to\infty}a_n=1/2$ by definition

Question

Suppose that $$ a_n=\frac{n^2-2n+1}{2n^2+4n-1}. $$ I am trying to show that for each positive number $\epsilon$, there is a number $N$ such that $|a_n-L|<\epsilon$ whenever $n>N$.

Attempt

Clearly $L=1/2$ as $n\to\infty$. So, \begin{align} |a_n-L|&=\left|\frac{n^2-2n+1}{2n^2+4n-1}-\frac{1}{2}\right| \\ &=\left|\frac{-4n+3/2}{2n^2+4n-1}\right| \\ &\leq \left|\frac{-4n+3/2}{4n-1}\right| \\ &\leq \frac{4n}{4n-1}+\frac{3/2}{4n-1} \\ &= \frac{4n-1}{4n-1}+\frac{5/2}{4n-1} \\ &= 1+\frac{5/2}{4n-1}. \\ \end{align} We require $$ 1+\frac{5/2}{4n-1}<\epsilon\implies n>\frac{1}{4}\left(\frac{5}{2(\epsilon-1)}+1\right)=N. $$

Note that I have made the assumption that $n>0$ since we are considering $n\to\infty$. Is this logic correct? Have I made any mistakes or assumptions?

Answer 1

For the fisrt inequality step,

Since $2n^2+4n-1 \ge n^2$ and $\left|-4n+3/2\right| \le 4n$, \begin{align} |a_n-L|&=\left|\frac{n^2-2n+1}{2n^2+4n-1}-\frac{1}{2}\right| \\ &=\left|\frac{-4n+3/2}{2n^2+4n-1}\right| \\ &\leq \left|\frac{4n}{n^2}\right| \\ &=\frac{4}{n} \end{align}

Answer 2

\begin{align} |a_n-L|&=\left|\frac{n^2-2n+1}{2n^2+4n-1}-\frac{1}{2}\right| \\ &=\left|\frac{-4n+3/2}{2n^2+4n-1}\right| \\ &\leq\frac{4n+3/2}{2n^2+4n-1} \\ &\leq \frac{8n}{2n^2}=\dfrac4n \\ \end{align} We require $$\dfrac4n<\epsilon\iff n>\frac4{\epsilon}\,.$$

Consequently,

for any $\,\epsilon>0\,$ there exists $\,N=\max\left\{\left\lfloor\dfrac4{\epsilon}\right\rfloor,1\right\}\in\Bbb N\,$ such that $\;|a_n-L|<\epsilon\;\;$ whenever $\,n>N\,.\\\\$

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Addendum :

$n>N\implies n>\dfrac4{\epsilon}\,.$

Indeed, if $\,n>N=\max\left\{\left\lfloor\dfrac4{\epsilon}\right\rfloor,1\right\}\,,\,$ then $\,n>\left\lfloor\dfrac4{\epsilon}\right\rfloor\,,\,$ hence, $\,n\geq\left\lfloor\dfrac4{\epsilon}\right\rfloor+1>\dfrac4{\epsilon}\,.$

Answer 3

You have to be careful with $ \epsilon -1$ as it can easily be negative if $\epsilon < 1$. So, your inequality for n is false. It's true if $\epsilon - 1 > 0$. The problem is that you deleted $2n^2$ and the inequality is weak and can't prove the theorem, because clearly $|\frac{-4n + \frac{3}{2}}{4n-1}|$ approaches 1, not $0$. As a hint, in that very step divide the numerator and the denominator by $n^2$, and use Archimedean property of real numbers: $\forall \epsilon >0, \exists \ n_0 \in N$ such that $ \frac{1}{n} < \epsilon$ when $ n > n_0$.