zadanie granica ciągu - pomocy!

[camilllla1]

\(\Lim_{n\to\infty}\frac{4+8+12+⋯+4n}{4n+2}\) oblicz granicę ciągu

[eresh]

\(\Lim_{n\to\infty}\frac{4+8+12+⋯+4n}{4n+2}\) oblicz granicę ciągu

\(\Lim_{n\to\infty}\frac{4(1+2+3+...+n)}{4n+2}=\Lim_{n\to\infty}\frac{4\cdot\frac{1+n}{2}\cdot n}{4n+2}=\\=\Lim_{n\to\infty}\frac{2n+2n^2}{4n+2}=\Lim_{n\to\infty}\frac{\frac{2}{n}+2}{\frac{4}{n}+\frac{2}{n^2}}=[\frac{2}{0}]=\infty\)

[camilllla1]

DZIĘKUJĘ! <3

[Jerry]

\(\cdots =\Lim_{n\to\infty}\frac{2n+2n^2}{4n+2}=\Lim_{n\to\infty}\frac{\frac{2}{n}+2}{\frac{4}{n}+\frac{2}{n^2}}=[\frac{2}{0}]=\infty\)

Wg mnie lepiej wygląda
\(\Lim_{n\to\infty}\frac{2n+2n^2}{4n+2}=\Lim_{n\to\infty}{n^2\over n}\cdot \frac{{2\over n}+2}{4+{2\over n}}=\infty\cdot{0+2\over4+0}=+\infty\)
bo symbol \(\left[\frac{2}{0}\right]\) wymaga wyjaśnienia \(\left[\frac{2}{0^\color{red}{+}}\right]=\color{red}{+}\infty\) albo \(\left[\frac{2}{0^\color{red}{-}}\right]=\color{red}{-}\infty\)

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