Granice ciągów (2)

[lubiepaczki]

Obliczyć granice ciągów:
\(f) \lim_{n\to \infty} \frac{log^{3}n^{2}}{log^{2}n^{3}}
g) \lim_{n\to \infty} \frac{lg_2 n}{log_4 n}
h) \lim_{n\to \infty} \frac{ lg_n 2}{lg_3 n}
i) \lim_{n\to \infty} \frac{n!}{ \left( n+1\right)! - n! }
j) \lim_{n\to \infty} \frac{ \left( \sqrt{n^{2}+1} +n \right)^{2} }{ \sqrt[]{n^{3}+1} }\)

[arqivus]

\(i) \lim_{n\to \infty} \frac{n!}{(n+1)!+n!}= \lim_{n\to \infty} \frac{n!}{(n)!\cdot (n+1)+n!}= \lim_{n\to \infty} \frac{1}{(n+1)+1}=0\)

[Galen]

\(f)\\ \lim_{n\to \infty } \frac{(logn^2)^3}{(logn^3)^2}= \lim_{n\to \infty } \frac{(2logn)^3}{(3logn)^2}= \lim_{n\to \infty } \frac{8 \cdot (logn)^3}{9(logn)^2}= \lim_{n\to \infty } \frac{8logn}{9}=+ \infty \\
g)\\
\lim_{n\to \infty } \frac{log_2n}{ \frac{log_2n}{log_24} }= \lim_{n\to \infty }log_2n \cdot \frac{2}{log_2n}=2\)

[Galen]

h)
\(\frac{lg_n2}{lg_3n}=\frac{\frac{lg_32}{lg_3n}}{lg_3n}= \frac{lg_32}{lg_3n} \cdot \frac{1}{lg_3n}= \frac{lg_32}{(lg_3n)^2}\\
granica:\\
= \frac{lg_32}{+ \infty }=0\)