7.

Limes oblika: ∞∞ \frac{\infin}{\infin}

TEKST ZADATKA

Odrediti graničnu vrednost:

lim⁡x→∞3x2+1x3+1\lim_{{x} \to {\infin}} \frac{3x^2+1}{x^3 + 1}
REŠENJE ZADATKA

Uvrstiti graničnu vrednost u izraz.

lim⁡x→∞3∞2+1∞3+1=lim⁡x→∞3∞+1∞+1\lim_{{x} \to {\infin}} \frac{3\infin^2+1}{\infin^3 + 1} = \lim_{{x} \to {\infin}} \frac{3\infin+1}{\infin + 1}

Granična vrednost je neodređenog oblika ∞∞. \frac{\infin}{\infin} .

lim⁡x→∞∞+1∞+1=lim⁡x→∞∞∞\lim_{{x} \to {\infin}} \frac{\infin+1}{\infin + 1} = \lim_{{x} \to {\infin}} \frac{\infin}{\infin}

Imenilac i brojilac se dele najvećim stepenom.

lim⁡x→∞3x2+1x3◯+1=lim⁡x→∞3x2x3+1x3x3x3+1x3\lim_{{x} \to {\infin}} \frac{3x^2+1}{x^{\textcircled{3}} + 1} = \lim_{{x} \to {\infin}} \frac{\frac{3x^2}{x^3}+\frac{1}{x^3}}{\frac{x^3}{x^3} + \frac{1}{x^3}}

Srediti izraz.

lim⁡x→∞3x2x31+1x3x3x31+1x3=lim⁡x→∞3x+1x31+1x3\lim_{{x} \to {\infin}} \frac{\frac{3\cancel{x^2}}{x^{\cancel{3}1}}+\frac{1}{x^3}}{\cancel{\frac{x^3}{x^3}}1 + \frac{1}{x^3}} = \lim_{{x} \to {\infin}} \frac{\frac{3}{x}+\frac{1}{x^3}}{1 + \frac{1}{x^3}}

Uvrstiti x→∞. x \to \infin .

lim⁡x→∞3∞+1∞31+1∞3=lim⁡x→∞3∞+1∞1+1∞\lim_{{x} \to {\infin}} \frac{\frac{3}{\infin}+\frac{1}{\infin^3}}{1 + \frac{1}{\infin^3}} = \lim_{{x} \to {\infin}} \frac{\frac{3}{\infin}+\frac{1}{\infin}}{1 + \frac{1}{\infin}}
lim⁡x→∞3∞0+1∞01+1∞0=lim⁡x→∞0+01+0\lim_{{x} \to {\infin}} \frac{\cancel{\frac{3}{\infin}}0+\cancel{\frac{1}{\infin}}0}{1 + \cancel{\frac{1}{\infin}}0} = \lim_{{x} \to {\infin}} \frac{0+0}{1 +0}

Srediti izraz.

lim⁡x→∞3∞0+1∞01+1∞0=lim⁡x→∞0+01+0\lim_{{x} \to {\infin}} \frac{\cancel{\frac{3}{\infin}}0+\cancel{\frac{1}{\infin}}0}{1 + \cancel{\frac{1}{\infin}}0} = \lim_{{x} \to {\infin}} \frac{0+0}{1 +0}
lim⁡x→∞01=lim⁡x→∞0=0\lim_{{x} \to {\infin}} \frac{0}{1} = \lim_{{x} \to {\infin}} 0 = 0

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