Teraz z załącznikiem , przepraszam , proszę o pomoc w rozwiązaniu
A) tam gdzie mamy n w mianowniku a w liczniku liczbe, to te rzeczy dążą do zera przy n dążacym do nieszkończoności
ponieważ w liczniku mamy n a w mianowniku tylko 4
B)
C)
Sytuacja odwrotna niż w A), n jest w mianowniku więc całośc dąży do 0
D)
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A)
tam gdzie mamy n w mianowniku a w liczniku liczbe, to te rzeczy dążą do zera przy n dążacym do nieszkończoności
ponieważ w liczniku mamy n a w mianowniku tylko 4
B)![\lim_{n \to \infty} \frac{3n+1}{1-5n} =\lim_{n \to \infty} \frac{n(3+\frac{1}{n})}{n(\frac{1}{n}-5)} = \lim_{n \to \infty} \frac{(3+\frac{1}{n})}{(\frac{1}{n}-5)} = -\frac{3}{5} \lim_{n \to \infty} \frac{3n+1}{1-5n} =\lim_{n \to \infty} \frac{n(3+\frac{1}{n})}{n(\frac{1}{n}-5)} = \lim_{n \to \infty} \frac{(3+\frac{1}{n})}{(\frac{1}{n}-5)} = -\frac{3}{5}](https://tex.z-dn.net/?f=+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7B3n%2B1%7D%7B1-5n%7D+%3D%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bn%283%2B%5Cfrac%7B1%7D%7Bn%7D%29%7D%7Bn%28%5Cfrac%7B1%7D%7Bn%7D-5%29%7D+%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7B%283%2B%5Cfrac%7B1%7D%7Bn%7D%29%7D%7B%28%5Cfrac%7B1%7D%7Bn%7D-5%29%7D+%3D+-%5Cfrac%7B3%7D%7B5%7D)
C)![\lim_{n \to \infty} \frac{4n+8}{n^2 -1} = \lim_{n \to \infty} \frac{n(4+\frac{8}{n})}{n(n -\frac{1}{n})} = \lim_{n \to \infty} \frac{(4+\frac{8}{n})}{(n -\frac{1}{n})} =0 \lim_{n \to \infty} \frac{4n+8}{n^2 -1} = \lim_{n \to \infty} \frac{n(4+\frac{8}{n})}{n(n -\frac{1}{n})} = \lim_{n \to \infty} \frac{(4+\frac{8}{n})}{(n -\frac{1}{n})} =0](https://tex.z-dn.net/?f=+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7B4n%2B8%7D%7Bn%5E2+-1%7D+%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bn%284%2B%5Cfrac%7B8%7D%7Bn%7D%29%7D%7Bn%28n+-%5Cfrac%7B1%7D%7Bn%7D%29%7D+%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7B%284%2B%5Cfrac%7B8%7D%7Bn%7D%29%7D%7B%28n+-%5Cfrac%7B1%7D%7Bn%7D%29%7D+%3D0)
Sytuacja odwrotna niż w A), n jest w mianowniku więc całośc dąży do 0
D)![\lim_{n \to \infty} \frac{5n^2+5}{10n^2-4} = \lim_{n \to \infty} \frac{n^2(5n+\frac{5}{n^2})}{n^2(10-\frac{4}{n^2})} = \\ \lim_{n \to \infty} \frac{(5n+\frac{5}{n^2})}{(10-\frac{4}{n^2})} = \frac{5}{10}= \frac{1}{2} \lim_{n \to \infty} \frac{5n^2+5}{10n^2-4} = \lim_{n \to \infty} \frac{n^2(5n+\frac{5}{n^2})}{n^2(10-\frac{4}{n^2})} = \\ \lim_{n \to \infty} \frac{(5n+\frac{5}{n^2})}{(10-\frac{4}{n^2})} = \frac{5}{10}= \frac{1}{2}](https://tex.z-dn.net/?f=+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7B5n%5E2%2B5%7D%7B10n%5E2-4%7D+%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bn%5E2%285n%2B%5Cfrac%7B5%7D%7Bn%5E2%7D%29%7D%7Bn%5E2%2810-%5Cfrac%7B4%7D%7Bn%5E2%7D%29%7D+%3D+%5C%5C+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7B%285n%2B%5Cfrac%7B5%7D%7Bn%5E2%7D%29%7D%7B%2810-%5Cfrac%7B4%7D%7Bn%5E2%7D%29%7D+%3D+%5Cfrac%7B5%7D%7B10%7D%3D+%5Cfrac%7B1%7D%7B2%7D)