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3 granice funkcji
: 7 maja 2009, o 09:34
autor: askasid
Proszę o pomoc w policzeniu takich granic:
1.\(\displaystyle{ \lim_{x \to 0^{+} } ( \sqrt{x} )^{x}}\)
2.\(\displaystyle{ \lim_{x\to 0^{+} }( \frac{1}{e ^{x}-1 }- \frac{1}{x} )}\)
3.\(\displaystyle{ \lim_{x\to 0^{+}}x ^{ \sqrt{x} }}\)
3 granice funkcji
: 7 maja 2009, o 10:00
autor: scyth
1.
\(\displaystyle{ \left(\sqrt{x}\right)^x=\left(x^{\frac{1}{2}}\right)^x=x^{\frac{x}{2}} = e^{\ln x^{\frac{x}{2}}} = e^{\frac{x}{2} \ln x} \\
\lim_{x \to 0^+} \frac{x}{2} \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac{2}{x}} = \left[ \frac{-\infty}{\infty} \right] \stackrel{H}{=} \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{2}{x^2}} = \lim_{x \to 0^+} - \frac{x}{2}} = 0 \\
\Rightarrow \lim_{x \to 0^+} \left(\sqrt{x}\right)^x = \lim_{x \to 0^+} e^{\frac{x}{2} \ln x} =
e^{\lim_{x \to 0^+} \frac{x}{2} \ln x} = e^0 = 1}\)-- 7 maja 2009, 10:04 --2.
\(\displaystyle{ \lim_{x \to 0^+} \left(\frac{1}{e^x-1} - \frac{1}{x} \right) =
\lim_{x \to 0^+} \frac{x-e^x+1}{x(e^x-1)} = \left[ \frac{0}{0} \right] \stackrel{H}{=}
\lim_{x \to 0^+} \frac{1-e^x}{e^x-1+xe^x} =\\= \left[ \frac{0}{0} \right] \stackrel{H}{=}
\lim_{x \to 0^+} \frac{-e^x}{e^x+e^x+xe^x} = \frac{-1}{1+1+0} = - \frac{1}{2}}\)
3. tak jak 1.