oblicz
: 15 wrz 2007, o 21:52
oblicz:
\(\displaystyle{ \lim_{n\to }\frac{1+2+3+...+n}{\sqrt{9n^4+1}}}\)
\(\displaystyle{ \lim_{n\to }\frac{1+2+3+...+n}{\sqrt{9n^4+1}}}\)
\(\displaystyle{ \lim_{n\to }\frac{1+2+3+...+n}{\sqrt{9n^4+1}}=
\lim_{n\to }\frac{\frac{(1+n)n}{2}}{\sqrt{9n^4+1}}=
\lim_{n\to }\frac{n^2+n}{2\sqrt{9n^4+1}}=
\lim_{n\to }\frac{1+\frac{1}{n}}{2\sqrt{9+\frac{1}{n^4}}}=
\frac{1}{6}}\)
POZDRO