[tex]\lim_{n \to \infty} \left(\frac{3n}{3n-5}\right)^{4n}=\lim_{n \to \infty} \left(\frac{3n-5+5}{3n-5}\right)^{4n}=\lim_{n \to \infty} \left(1+\frac{5}{3n-5}\right)^{4n}=\lim_{n \to \infty} \left(1+\frac{1}{\frac{3n-5}{5}}\right)^{4n}=\lim_{n \to \infty} \left(\left(1+\frac{1}{\frac{3n-5}{5}}\right)^{\frac{3n-5}{5}}\right)^{\frac{5}{3n-5}*4n}=\lim_{n \to \infty} \left(\left(1+\frac{1}{\frac{3n-5}{5}}\right)^{\frac{3n-5}{5}}\right)^{\frac{20n}{3n-5}}=[/tex][tex]\lim_{n \to \infty} \left(\left(1+\frac{1}{\frac{3n-5}{5}}\right)^{\frac{3n-5}{5}}\right)^{\frac{20}{3-\frac{5}{n}}}=e^\frac{20}{3}[/tex]
[tex]\lim_{n \to \infty} \left(\sqrt{16n^2+16n}-4n\right)=\lim_{n \to \infty} \left(\sqrt{16n^2+16n}-4n\right)*\frac{\sqrt{16n^2+16n}+4n}{\sqrt{16n^2+16n}+4n}=\lim_{n \to \infty} \frac{16n^2+16n-16n^2}{\sqrt{16n^2+16n}+4n}=\lim_{n \to \infty} \frac{16n}{\sqrt{16n^2+16n}+4n}=\lim_{n \to \infty} \frac{16n}{n\left(\sqrt{16+\frac{16}{n}}+4\right)}=\lim_{n \to \infty} \frac{16}{\sqrt{16+\frac{16}{n}}+4}=\frac{16}{4+4}=2[/tex]
[tex] \lim_{x \to 0} \frac{5x-3x}{25x-9x}= \lim_{x \to 0} \frac{2x}{16x}=\lim_{x \to 0} \frac{2}{16}=\frac{1}{8}[/tex]
Czy taka była granica wyjściowa, bo nie wiem, czy dobrze odczytałem. Jeśli nie, to proszę dać znać, to poprawię.
