wyznacz granice funkcji
: 30 lis 2012, o 18:27
a)\(\displaystyle{ \lim_{ n\to \infty } = \left( 0,\underbrace{9\ldots9}_{n} \right) ^{10n}}\)
b)\(\displaystyle{ \lim_{n \to \infty } =\sin \sqrt{n+1} -\sin \sqrt{n}}\)
c)\(\displaystyle{ \lim_{n \to \infty }=n \left( \sqrt[n]{3} -1 \right)}\)
d)\(\displaystyle{ \lim_{n \to \infty } = \frac {\sin ^{2}n +4n }{3n-1}}\)
e)\(\displaystyle{ \lim_{ n\to \infty } =n \left( \frac{1}{1+n ^{2} } + ... + \frac{1}{n+n ^{2} } \right)}\)
f) \(\displaystyle{ \lim_{ n\to \infty } =n \left( \frac{1}{1+n ^{2} } + \frac{2}{2+n ^{2} } + ... + \frac{n}{n+n ^{2} } \right)}\)
g) \(\displaystyle{ \lim_{ n\to \infty } =\sin \left( \pi \sqrt{n ^{2}+n } \right)}\)
b)\(\displaystyle{ \lim_{n \to \infty } =\sin \sqrt{n+1} -\sin \sqrt{n}}\)
c)\(\displaystyle{ \lim_{n \to \infty }=n \left( \sqrt[n]{3} -1 \right)}\)
d)\(\displaystyle{ \lim_{n \to \infty } = \frac {\sin ^{2}n +4n }{3n-1}}\)
e)\(\displaystyle{ \lim_{ n\to \infty } =n \left( \frac{1}{1+n ^{2} } + ... + \frac{1}{n+n ^{2} } \right)}\)
f) \(\displaystyle{ \lim_{ n\to \infty } =n \left( \frac{1}{1+n ^{2} } + \frac{2}{2+n ^{2} } + ... + \frac{n}{n+n ^{2} } \right)}\)
g) \(\displaystyle{ \lim_{ n\to \infty } =\sin \left( \pi \sqrt{n ^{2}+n } \right)}\)