O limita
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Mateescu Constantin
- Newton
- Posts: 307
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O limita
Calculati : \( \lim_{x\to\infty}\ \left(\ 2^{\frac {2-x}x}\ +\ 4^{\frac {4-x}x}\ +\ 6^{\frac {6-x}x}\ +\ 12^{\frac {12-x}x}\ \right)^x \) .
Last edited by Mateescu Constantin on Sun Aug 29, 2010 4:06 pm, edited 1 time in total.
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DrAGos Calinescu
- Thales
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Expresia \( 2^{\frac{2-x}{x}}+4^{\frac{4-x}{x}}+6^{\frac{6-x}{x}}+12^{\frac{12-x}{x}}\to \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=1 \)
Deci \( \lim_{x\to\infty}\ \left( 2^{\frac{2-x}{x}}+4^{\frac{4-x}{x}}+6^{\frac{6-x}{x}}+12^{\frac{12-x}{x}}\ \right)^x=e^{\lim_{x\to\infty} x\ \left( 2^{\frac{2-x}{x}}+4^{\frac{4-x}{x}}+6^{\frac{6-x}{x}}+12^{\frac{12-x}{x}}-1\ \right)}=e^{\lim_{x\to\infty} x\ \left( 2^{\frac{2-x}{x}}-\frac 12+4^{\frac{4-x}{x}}-\frac{1}{4}+6^{\frac{6-x}{x}}-\frac{1}{6}+12^{\frac{12-x}{x}}-\frac{1}{12}\ \right)=e^{\lim_{x\to\infty} \ \left( \frac{2^{\frac{2}{x}}-1}{\frac{2}{x}}+\frac{4^{\frac{4}{x}}-1}{\frac{4}{x}}+\frac{6^{\frac{6}{x}}-1}{\frac{6}{x}}+\frac{12^{\frac{12}{x}}-1}{\frac{12}{x}}\ \right)}=e^{\ln 2+\ln 4+\ln 6+\ln 12}=e^{\ln 576}=576 \)
Deci \( \lim_{x\to\infty}\ \left( 2^{\frac{2-x}{x}}+4^{\frac{4-x}{x}}+6^{\frac{6-x}{x}}+12^{\frac{12-x}{x}}\ \right)^x=e^{\lim_{x\to\infty} x\ \left( 2^{\frac{2-x}{x}}+4^{\frac{4-x}{x}}+6^{\frac{6-x}{x}}+12^{\frac{12-x}{x}}-1\ \right)}=e^{\lim_{x\to\infty} x\ \left( 2^{\frac{2-x}{x}}-\frac 12+4^{\frac{4-x}{x}}-\frac{1}{4}+6^{\frac{6-x}{x}}-\frac{1}{6}+12^{\frac{12-x}{x}}-\frac{1}{12}\ \right)=e^{\lim_{x\to\infty} \ \left( \frac{2^{\frac{2}{x}}-1}{\frac{2}{x}}+\frac{4^{\frac{4}{x}}-1}{\frac{4}{x}}+\frac{6^{\frac{6}{x}}-1}{\frac{6}{x}}+\frac{12^{\frac{12}{x}}-1}{\frac{12}{x}}\ \right)}=e^{\ln 2+\ln 4+\ln 6+\ln 12}=e^{\ln 576}=576 \)