Úlohy - limita - úlohy s goniometrickými funkcemi
Úloha
\(\lim_{x \to 0} \sin x\)
\(\lim_{x \to \frac {\pi}{2}} \sin x\)
\(\lim_{x \to \pi} \sin x\)
- \(\lim_{x \to 0} \sin x = 0\)
\(\lim_{x \to \frac {\pi}{2}} \sin x = 1\)
\(\lim_{x \to \pi} \sin x = 0\)
Úloha
\(\lim_{x \to 0} \frac {\cos x - 1}{\sin^2 x} = \)
- \(= \lim_{x \to 0} \frac {\cos x - 1}{\sin^2 x} \cdot \frac {\cos x + 1}{\cos x + 1} = \)
- \(= \lim_{x \to 0} \frac {(-1)(1 - \cos^2 x)}{(1 - \cos^2 x)(\cos x + 1)} = \)
- \(= \lim_{x \to 0} \frac {-1}{\cos x + 1} = \)
- \(= - \frac {1}{2} \)
Úloha
\(\lim_{x \to 0} \frac {\sin 3x}{5x} = \)
- \(= \lim_{x \to 0} \frac {\sin 3x}{5x} \cdot \frac {3}{3} = \)
- \(= \lim_{x \to 0} \frac {\sin 3x}{3x} \cdot \frac {3}{5} = \)
- \(= \frac {3}{5} \)
Úloha
\(\lim_{x \to 0} \frac {\mathrm{tg} 5x}{3x} = \)
- \(= \lim_{x \to 0} \frac {\sin5x}{3x\cos5x} \cdot \frac {5}{5} = \)
- \(= \lim_{x \to 0} \frac {\sin 5x}{5x} \cdot \frac {5}{3\cos5x} = \)
- \(= \frac {5}{3} \)
Úloha
\(\lim_{x \to 0} \frac {1 - \cos2x}{x\sin x} = \)
- \(= \lim_{x \to 0} \frac {1 - (\cos^2 x - \sin^2 x)}{x \sin x} = \)
- \(= \lim_{x \to 0} \frac {2 \sin^2 x}{x \sin x} = \)
- \(= \lim_{x \to 0} \frac {2 \sin x}{x} = \)
- \(= 2 \)
Úloha
\(\lim_{x \to 0} \frac {1 - \cos2x + \mathrm{tg}^{2}x}{x\sin x} = \)
- \(= \lim_{x \to 0} \frac {2 \sin^2x + \frac{\sin^2 x}{\cos^2 x}}{x\sin x} = \)
- \(= \lim_{x \to 0} \frac {\sin^2 x}{x\sin x}\left(2 + \frac{1}{\cos^2 x}\right) = \)
- \(= 3 \)
Úloha
\(\lim_{x \to 0} \frac {\mathrm{tg} x - \sin x}{x^3} = \)
- \(= \lim_{x \to 0} \frac {\frac{\sin x}{\cos x} - \sin x}{x^3} = \)
- \(= \lim_{x \to 0} \frac {\sin x}{x^3} \cdot \frac{1 - \cos x}{\cos x} \cdot \frac {1 + \cos x}{1 + \cos x} = \)
- \(= \lim_{x \to 0} \frac {\sin^3 x}{x^3} \cdot \frac{1}{\cos x (1 + \cos x)} = \)
- \(= \frac {1}{2} \)
Úloha
\(\lim_{x \to 0} \frac {\sin 2x}{\sqrt{x + 2} - \sqrt{2}} = \)
- \(= \lim_{x \to 0} \frac {\sin 2x}{\sqrt{x + 2} - \sqrt{2}} \cdot \frac{(\sqrt{x + 2} + \sqrt{2})}{(\sqrt{x + 2} + \sqrt{2})} = \)
- \(= \lim_{x \to 0} \frac {\sin 2x}{x} \cdot \left(\sqrt{x + 2} + \sqrt{2} \right ) \frac{2}{2} = \)
- \(= 4\sqrt{2} \)
