Granica funkcji
: 17 cze 2012, 20:00
\(\lim_{x \to 0}\(\frac{x}{\sin x}\)^{\frac{1}{\sin^2x}}\)
\(\lim_{x\to 0}\(\frac{x}{\sin x}\)^{\frac{1}{\sin^2x}}=\lim_{x\to 0}\(\frac{x}{\sin x}\)^{\frac{1}{\frac{x}{\sin x}-1} \cdot \(\frac{x}{\sin x}-1\) \cdot \frac{1}{\sin^2x}}=\lim_{x\to 0}\[\(\frac{x}{\sin x}\)^{\frac{1}{\frac{x}{\sin x}-1}}\]^{\frac{x-\sin x}{\sin^3x}}
\lim_{x\to 0}\frac{x-\sin x}{\sin^3x}=^H\lim_{x\to 0}\frac{1-\cos x}{3\sin^2x\cos x}=^H\lim_{x\to 0}\frac{\sin x}{6\sin x\cos^2x-3\sin^3x}=\lim_{x\to 0}\frac{1}{6\cos^2x-3\sin^2x}=\frac{1}{6}
\lim_{x\to 0}\(\frac{x}{\sin x}\)^{\frac{1}{\sin^2x}}=e^{\frac{1}{6}}\)