865.

Trigonometrijski limes

TEKST ZADATKA

Odrediti graničnu vrednost:

limx12(sin(2x1)4x21+8x312x2x)\lim_{{x} \to {\frac12}}\bigg(\frac{\sin(2x-1)}{4x^2-1}+\frac{8x^3-1}{2x^2-x}\bigg)

REŠENJE ZADATKA

Primeniti formulu za razliku kubova: a3b3=(ab)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)

limx12(sin(2x1)4x21+(2x1)(4x2+2x+1)x(2x1))\lim_{{x} \to {\frac12}}\bigg(\frac{\sin(2x-1)}{4x^2-1}+\frac{(2x-1)(4x^2+2x+1)}{x(2x-1)}\bigg)

Primeniti formulu za razliku kvadrata: a2b2=(ab)(a+b)a^2-b^2=(a-b)(a+b)

limx12(sin(2x1)(2x1)(2x+1)+(2x1)(4x2+2x+1)x(2x1))limx12(sin(2x1)(2x1)(2x+1)+4x2+2x+1x)\lim_{{x} \to {\frac12}}\bigg(\frac{\sin(2x-1)}{(2x-1)(2x+1)}+\frac{(2x-1)(4x^2+2x+1)}{x(2x-1)}\bigg) \\ \lim_{{x} \to {\frac12}}\bigg(\frac{\sin(2x-1)}{(2x-1)(2x+1)}+\frac{4x^2+2x+1}{x}\bigg)

Raščlaniti izraz.

limx12sin(2x1)2x1limx1212x+1+limx124x2+2x+1x\lim_{{x} \to {\frac12}}\frac{\sin(2x-1)}{2x-1}\cdot\lim_{{x} \to {\frac12}}\frac1{2x+1}+\lim_{{x} \to {\frac12}}\frac{4x^2+2x+1}{x}

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

1limx1212x+1+limx124x2+2x+1x1\cdot\lim_{{x} \to {\frac12}}\frac1{2x+1}+\lim_{{x} \to {\frac12}}\frac{4x^2+2x+1}{x}

Zameniti x=12.x=\frac12.

1212+1+4(12)2+212+112132\frac1{2\cdot\frac12+1}+\frac{4\cdot(\frac12)^2+2\cdot\frac12+1}{\frac12} \\ \frac{13}2

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