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Granice
: 3 wrz 2007, o 17:18
autor: jaczek
Mam prośbę o pomoc w obliczeniu trzech granic.
1. \(\displaystyle{ \lim_{-\infty}x(\sqrt{x^{2}+5}+x)}\)
2. \(\displaystyle{ \lim_{1}(1-x)\tan{\frac{\pi{x}}{2}}}\)
3. \(\displaystyle{ \lim_{\infty}({\tanh{x}})^{e^{2x}}}\)
Dzięki!
Granice
: 3 wrz 2007, o 17:34
autor: jasny
1.
\(\displaystyle{ =\lim_{x\to-\infty}\frac{x(\sqrt{x^2+5}+x)(\sqrt{x^2+5}-x)}{\sqrt{x^2+5}-x}= \lim_{x\to-\infty}\frac{5x}{\sqrt{x^2+5}-x}=
\lim_{x\to-\infty}\frac{5x}{-x\sqrt{1+\frac{5}{x^2}}-x}= \lim_{x\to-\infty}\frac{5}{-\sqrt{1+\frac{5}{x^2}}-1}=-\frac{5}{2}}\)
[ Dodano: 3 Września 2007, 19:05 ]
2.
\(\displaystyle{ =\lim_{x\to1}\frac{1-x}{\cot\frac{\pi x}{2}} =\lim_{x\to1}\frac{(1-x)'}{(\cot\frac{\pi x}{2})'}= \lim_{x\to1}\frac{-1}{-\frac{\pi}{2}\cdot\frac{1}{\sin^2\frac{\pi x}{2}}}= \lim_{x\to1}\frac{2\sin^2\frac{\pi x}{2}}{\pi}=\frac{2}{\pi}}\)
Granice
: 3 wrz 2007, o 20:31
autor: max
2. Bez Hospitala:
\(\displaystyle{ \lim_{x\to 1}\left((1 - x)\tan \frac{\pi}{2}x\right) \stackrel{t = 1 - x}{=}\lim_{t\to 0}\left(t\cdot \tan\left(\frac{\pi}{2} - \frac{\pi}{2}t\right)\right) = \\
= \lim_{t\to 0}\left(t\cdot \cot \frac{\pi}{2}t\right) = \lim_{t\to 0} \frac{\frac{\pi}{2}t\cos \frac{\pi}{2}t}{\frac{\pi}{2}\sin \frac{\pi}{2}t} = \frac{2}{\pi}}\)
3.
\(\displaystyle{ \lim_{x\to +\infty}(\tanh x)^{e^{2x}} = \lim_{x\to\infty}\left(\frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}\right)^{e^{2x}} = \lim_{x\to\infty}\left(1 + \frac{-2e^{-x}}{e^{x} + e^{-x}}\right)^{e^{2x}} = \\
= \lim_{x\to\infty}\left(\left(1 + \frac{2}{-e^{2x} - 1}\right)^{-e^{2x}-1}\cdot\left(1 + \frac{1}{-e^{2x} - 1}\right)\right)^{-1} = (e^{2}\cdot 1)^{-1} = \frac{1}{e^{2}}}\)