853.

Trigonometrijski limes

TEKST ZADATKA

Odrediti graničnu vrednost:

limx01cosxx2\lim_{{x} \to {0}}\frac{1-\cos x}{x^2}

REŠENJE ZADATKA

Primeniti formulu za sinus poluugla: sin2α2=1cosα2\sin^2{\frac {\alpha} 2}=\frac {1-\cos{\alpha}} 2

limx02sin2x2x2\lim_{{x} \to {0}}\frac{2\sin^2\frac x2}{x^2}

Raščlaniti izraz.

limx02limx0sinx2xlimx0sinx2x\lim_{{x} \to {0}}2\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{x}\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{x}

Pomnožiti imenioce sa 22.\frac22.

2limx0sinx2x22limx0sinx2x222limx0sinx2x2limx012limx0sinx2x2limx0122limx0sinx2x212limx0sinx2x21212limx0sinx2x2limx0sinx2x22\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{x\cdot\frac22}\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{x\cdot\frac22}\\ 2\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{\frac x2}\cdot\lim_{{x} \to {0}}\frac12\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{\frac x2}\cdot\lim_{{x} \to {0}}\frac12 \\ 2\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{\frac x2}\cdot\frac12\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{\frac x2}\cdot\frac12 \\ \frac12\lim_{{x} \to {0}}\frac{\sin\frac x2}{\frac x2}\cdot\lim_{{x} \to {0}}\frac{\sin\frac x2}{\frac x2}

Primeniti tablični limes: limx0sinxx=1 \lim_{{x} \to {0}}\frac {\sin{x}} {x}=1

121112\frac12\cdot1\cdot1 \\ \frac12

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