Granica funkcji

[pavi_123]

Oblicz granicę \(\Lim_{x\to\infty}\left(\frac{x}{\arctg{x}}-\frac{2x}{\pi}\right)\).

[radagast]

Oblicz granicę \(lim_{x\to\infty}\frac{x}{\arctg{x}}-\frac{2x}{\pi}\).

\(\Lim_{x\to\infty}(\frac{x}{\arctg{x}}-\frac{2x}{\pi})=\Lim_{x\to\infty}x(\frac{1}{\arctg{x}}-\frac{2}{\pi})=\Lim_{x\to\infty} \frac{\frac{1}{\arctg{x}}-\frac{2}{\pi} }{ \frac{1}{x} }\nad{H}{=}\Lim_{x\to\infty} \frac{-\frac{1}{\arctg^2{x}} \frac{1}{1+x^2} }{ -\frac{1}{x^2} } =\\ \quad=\Lim_{x\to\infty} \frac{1}{\arctg^2{x}} \frac{x^2}{1+x^2}={1\over( \frac{\pi}{2})^2}= \frac{4}{\pi^2} \)
bez de l'Hospitala pewnie się da ale mi nie chce wyjść )

[Jerry]

bez de l'Hospitala pewnie się da ...

Niech
\({\pi\over2}-t=\arctg x\wedge t\to0\)
wtedy
\(x={1\over\tg t}\)
i
\(\Lim_{x\to\infty}\left(\frac{x}{\arctg{x}}-\frac{2x}{\pi}\right)=
\Lim_{t\to0}\left(\frac{1}{\tg t({\pi\over2}-t)}-{2\over\pi\tg t}\right)=
\Lim_{t\to0}{t\over\tg t}\cdot{2\over\pi({\pi\over2}-t)}=1\cdot{2\over\pi\cdot{\pi\over2}}={4\over\pi^2}\)

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