862.

Trigonometrijski limes

TEKST ZADATKA

Odrediti graničnu vrednost:

lim⁡x→02x2cos⁡8x−cos⁡2x\lim_{{x} \to {0}}\frac{2x^2}{\cos8x-\cos2x}
REŠENJE ZADATKA

Primeniti formulu za razliku kosinusa: cos⁡α−cos⁡β=−2sin⁡α+β2sin⁡α−β2 \cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}2\sin\frac{\alpha-\beta}2

lim⁡x→02x2−2sin⁡8x+2x2sin⁡8x−2x2lim⁡x→0x2−sin⁡5xsin⁡3x\lim_{{x} \to {0}}\frac{2x^2}{-2\sin\frac{8x+2x}2\sin\frac{8x-2x}2} \\ \lim_{{x} \to {0}}\frac{x^2}{-\sin5x\sin3x}

Raščlaniti izraz.

−lim⁡x→0xsin⁡5x⋅lim⁡x→0xsin⁡3x-\lim_{{x} \to {0}}\frac{x}{\sin5x}\cdot\lim_{{x} \to {0}}\frac x{\sin3x}

Preoblikovati izraz pod limesom radi lakše primene tabličnog limesa.

−lim⁡x→01sin⁡5xx⋅lim⁡x→01sin⁡3xx-\lim_{{x} \to {0}}\frac1{\frac{\sin5x}x}\cdot\lim_{{x} \to {0}}\frac 1{\frac{\sin3x}x}

Pomnožiti imenilac sa 55,\frac 55, odnosno sa 33.\frac 33.

−lim⁡x→01sin⁡5xx⋅55⋅lim⁡x→01sin⁡3xx⋅33−lim⁡x→01sin⁡5x5x⋅5⋅lim⁡x→01sin⁡3x3x⋅3-\lim_{{x} \to {0}}\frac1{\frac{\sin5x}x\cdot\frac55}\cdot\lim_{{x} \to {0}}\frac 1{\frac{\sin3x}x\cdot\frac33} \\ -\lim_{{x} \to {0}}\frac1{\frac{\sin5x}{5x}\cdot5}\cdot\lim_{{x} \to {0}}\frac 1{\frac{\sin3x}{3x}\cdot3}

Primeniti tablični limes: lim⁡x→0sin⁡xx=1 \lim_{{x} \to {0}}\frac {\sin{x}} {x}=1

−15⋅13−115-\frac15\cdot\frac13 \\ -\frac1{15}

Balkan Tutor Sva Prava Zadržana © 2025

Politika privatnosti