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Granica
: 28 lut 2017, 11:43
autor: mela1015
Oblicz granicę
a) \(\Lim_{x\to 0 } \frac{xctgx-1}{x^2}\)
: 28 lut 2017, 11:53
autor: eresh
\(\Lim_{x\to 0}\frac{x\ctg x-1}{x^2}=\Lim_{x\to 0}\frac{\frac{x\cos x}{\sin x}-1}{x^2}=\Lim_{x\to 0}\frac{x\cos x-\sin x}{x^2\sin x}= \left[ \frac{0}{0}\right]=\\=\Lim_{x\to 0}\frac{\cos x-x\sin x-\cos x}{2x\sin x+x^2\cos x}=\Lim_{x\to 0}\frac{-\sin x}{2\sin x+x\cos x}= \left[\frac{0}{0} \right]=\Lim_{x\to 0}\frac{-\cos x}{2\cos x+\cos x-x\sin x}=\frac{-1}{2+1-0}=-\frac{1}{3}\)