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GRANICA CIĄGU
: 18 lis 2016, 12:55
autor: anonymous2
Oblicz granice ciągów:
a) \(\Lim_{x\to \infty } \sqrt{2n^2-2n+1} - \sqrt{2n^2+n+2}\)
b) an=\(\sqrt[2n]{4^n+3^n+4^{2n}}\)
: 18 lis 2016, 13:09
autor: radagast
\(\Lim_{n\to \infty } \sqrt{2n^2-2n+1} - \sqrt{2n^2+n+2}=\\
\Lim_{n\to \infty } \left(\sqrt{2n^2-2n+1} - \sqrt{2n^2+n+2} \right) \cdot \frac{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} }{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} } =\\
\Lim_{n\to \infty } \frac{2n^2-2n+1- 2n^2-n-2 }{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} } =\\
\Lim_{n\to \infty } \frac{-3n-1 }{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} } =\\
\Lim_{n\to \infty } \frac{-3- \frac{1}{n} }{\sqrt{2- \frac{2}{n} +\frac{1}{n^2}} + \sqrt{2+\frac{1}{n}+\frac{2}{n^2}} } =- \frac{3}{2 \sqrt{2} }= \frac{3}{4} \sqrt{2}\)
: 18 lis 2016, 13:14
autor: radagast
\(\Lim_{n\to \infty } \sqrt[2n]{4^n+3^n+4^{2n}}= \Lim_{n\to \infty } \sqrt[2n]{4^{2n}} \cdot \frac{\sqrt[2n]{4^n+3^n+4^{2n}}}{ \sqrt[2n]{4^{2n}}}=\Lim_{n\to \infty } 4 \cdot \sqrt[2n]{ \frac{4^n}{4^{2n}} +\frac{3^n}{4^{2n}}+1}=4\)
: 18 lis 2016, 13:37
autor: anonymous2
radagast pisze:\(\Lim_{n\to \infty } \sqrt{2n^2-2n+1} - \sqrt{2n^2+n+2}=\\
\Lim_{n\to \infty } \left(\sqrt{2n^2-2n+1} - \sqrt{2n^2+n+2} \right) \cdot \frac{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} }{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} } =\\
\Lim_{n\to \infty } \frac{2n^2-2n+1- 2n^2-n-2 }{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} } =\\
\Lim_{n\to \infty } \frac{-3n-1 }{\sqrt{2n^2-2n+1} + \sqrt{2n^2+n+2} } =\\
\Lim_{n\to \infty } \frac{-3- \frac{1}{n} }{\sqrt{2- \frac{2}{n} +\frac{1}{n^2}} + \sqrt{2+\frac{1}{n}+\frac{2}{n^2}} } =- \frac{3}{2 \sqrt{2} }= \frac{3}{4} \sqrt{2}\)
W pierwszym przykładzie wyjdzie
\(-\frac{3 \sqrt{2} }{4}\) zgadza się ?