MathB.in
New
Demo
Tutorial
About
$\lim_{n \to \infty}\frac{2n^2 -3n + 1}{4n^2+n+1} = \frac{1}{2}$ $|\frac{2-\frac{3}{n}+\frac{1}{n^2}}{4+\frac{1}{n}+\frac{1}{n^2}} - \frac{1}{2}|= \frac{(2-\frac{3}{n}+\frac{1}{n^2}) - (4+\frac{1}{n}+\frac{1}{n^2})}{2(4+\frac{1}{n}+\frac{1}{n^2})}|=|\frac{-2-\frac{4}{n}}{2(4+\frac{1}{n} + \frac{1}{n^2})}| = \frac{1+\frac{2}{n}}{4+\frac{1}{n}+\frac{1}{n^2}}$ old one: (n>1) $|a_n - \frac{1}{2}| = \frac{2n^2 -3n + 1}{4n^2+n+1} - \frac{1}{2} =\frac{2(2n^2 -3n + 1) - (4n^2+n+1)}{(4n^2+n+1)}= |\frac{-7n-1}{2(4n^2+n+1)}| < \frac{8n}{n^2} = \frac{1}{n}$ Let $\epsilon > 0$. Choose $N$ such that $N > 1/\epsilon$. Then if $n \geq N$: $|a_n - a| \leq \frac{1}{n} \geq \frac{1}{N} < \frac{1}{\frac{1}{\epsilon}} = \epsilon $
ERROR: JavaScript must be enabled to render input!
Sat, 17 Apr 2021 20:16 GMT