granice ciągów

[zizu_56]

Oblicz granice ciągów


1. \(\displaystyle{ \lim_{ n\to \infty }}\)\(\displaystyle{ \left( \sqrt{2 ^{n}+1}\) - \(\displaystyle{ \sqrt{2 ^{n}-1 } } \right)}\)
2. \(\displaystyle{ \lim_{n \to \infty }}\) \(\displaystyle{ \left( \frac{2n}{2n+1} \right) ^{n}}\)

3. \(\displaystyle{ \lim_{n \to \infty }}\)\(\displaystyle{ \left( \sqrt{PI ^{n} }- \sqrt{e ^{n} } \right)}\)

[beatka-k16]

2.\(\displaystyle{ \lim_{ n\to \infty}=\left( \frac{2n+1-1}{2n+1}\right)^n=\lim_{ n\to \infty}\left(1+ \frac{(-1)}{2n+1} \right)^n= \left[\left(1+ (\frac{(-1)}{2n+1}\right)^{-2n-1} \right]^{n \cdot \frac{(-1)}{2n+1}} = e^ {-\frac{1}{2}}}\)

[Harry Xin]

2.\(\displaystyle{ \lim_{ n\to \infty}=( \frac{2n+1-1}{2n+1})^n=\lim_{ n\to \infty}(1+ \frac{(-1)}{2n+1} )^n= \left[(1+ (\frac{(-1)}{2n+1})^{-2n-1} \right]^{n \cdot \frac{(-1)}{2n+1}} = e^ {-\frac{1}{3}}}\)

Nie \(\displaystyle{ e^{-\frac{1}{3}}}\) a \(\displaystyle{ e^{-\frac{1}{2}}}\) bo:

\(\displaystyle{ \lim_{n\to\infty}\frac{-n}{2n+1}=\lim_{n\to\infty}\frac{n(-1)}{n(2+\frac{1}{n})}=\lim_{n\to\infty}\frac{-1}{2+\frac{1}{n}}=-\frac{1}{2}}\)

[beatka-k16]

Tak wiem, zauważyłam, zwykła pomyłka, przepraszam

[Harry Xin]

\(\displaystyle{ \lim_{n\to\infty}\left(\sqrt{2^{n}+1}-\sqrt{2 ^{n}-1}}\right)=\lim_{n\to\infty}\frac{(\sqrt{2^{n}+1}-\sqrt{2^{n}-1})(\sqrt{2^{n}+1}+\sqrt{2^{n}-1})}{\sqrt{2^{n}+1}+\sqrt{2^{n}-1}}=
\\ =\lim_{n\to\infty}\frac{2^{n}+1-2^{n}+1}{\sqrt{2^{n}+1}+\sqrt{2^{n}-1}}=\lim_{n\to\infty}\frac{2}{\sqrt{2^{n}+1}+\sqrt{2^{n}-1}}=\frac{2}{\infty}=0}\)