Ponovo ja sa razvojima, ponovo u nadi da ću razumeti bolje ovo čudo od o-a.
[dispmath]\boldsymbol{\lim_{n\to\infty}\Biggl(n\cdot\left(\frac{\sqrt[n]e+1}{\sqrt[n]e-1}-2\cdot n\right)\Biggr)=}\\
=\lim_{n\to\infty}\left(n\cdot\frac{e^{\frac{1}{n}}+1-2n\left(e^{\frac{1}{n}}-1\right)}{e^{\frac{1}{n}}-1}\right)\\
=\lim_{n\to\infty}\left(n\cdot\frac{1+\frac{1}{n}+\frac{1}{2n^2}+1-2n\cdot\biggl(1+\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{6n^3}+o\left(\frac{1}{n^3}\right)-1\biggr)}{\frac{1}{n}+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)}\right)\\
=\lim_{n\to\infty}\left(n\cdot\frac{2+\frac{1}{n}+\frac{1}{2n^2}-2-\frac{1}{n}-\frac{1}{3n^2}+o\left(\frac{1}{n^2}\right)}{\frac{1}{n}+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)}\right)\\
=\lim_{n\to\infty}\left(n\cdot\frac{\frac{1}{6}n+o\left(\frac{1}{n}\right)}{\frac{1}{n}+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)}\right)\\
=\lim_{n\to\infty}\left(\frac{\frac{1}{6}+o(1)}{1+o(1)}\right)\\
=\frac{1}{6}[/dispmath] Na osnovu čega da znam da razvijamo baš do trećeg?