158.

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

TEKST ZADATKA

Odrediti graničnu vrednost:

lim⁡x→+∞x2+1−x2+13x4+14−x4+15\lim_{{x} \to {+\infty}} \frac {\sqrt{x^2+1}-\sqrt[3]{x^2+1}} {\sqrt[4]{x^4+1}- \sqrt[5]{x^4+1}}
REŠENJE ZADATKA

Podeliti limes sa najvećim stepenom x,x, u ovom slučaju x1x^1 i u brojiocu i imeniocu.

lim⁡x→+∞x2+1x−x2+13xx4+14x−x4+15x\lim_{{x} \to {+\infty}} \frac {\frac{\sqrt{x^2+1}} {x}-\frac{\sqrt[3]{x^2+1}} {x}} {\frac{\sqrt[4]{x^4+1}} {x}- \frac{\sqrt[5]{x^4+1}} {x}}

Podvući xx ispod korena u svakom razlomku.

lim⁡x→+∞x2+1x2−x2+1x33x4+1x44−x4+1x55\lim_{{x} \to {+\infty}} \frac {\sqrt{\frac{x^2+1}{x^2}}-\sqrt[3]{\frac{x^2+1}{x^3}}} {\sqrt[4]{\frac{x^4+1}{x^4}}- \sqrt[5]{\frac{x^4+1}{x^5}}}

Srediti izraz.

lim⁡x→+∞1+1x2−1x+1x331+1x44−1x+1x55\lim_{{x} \to {+\infty}} \frac {\sqrt{1+\frac 1 {x^2}}-\sqrt[3]{\frac 1 x+\frac 1 {x^3}}} {\sqrt[4]{1+\frac 1 {x^4}}- \sqrt[5]{\frac 1 x+\frac 1 {x^5}}}
lim⁡x→+∞x2x2+1x2−x2x3+1x33x4x4+1x44−x4x5+1x55\lim_{{x} \to {+\infty}} \frac {\sqrt{\frac{x^2}{x^2}+\frac 1 {x^2}}-\sqrt[3]{\frac{x^2}{x^3}+\frac 1 {x^3}}} {\sqrt[4]{\frac{x^4}{x^4}+\frac 1 {x^4}}- \sqrt[5]{\frac{x^4}{x^5}+\frac 1 {x^5}}}

Uvrstiti x→∞. x \to \infin .

1+1∞2−1∞+1∞331+1∞44−1∞+1∞55=1+1∞2−1∞+1∞331+1∞44−1∞+1∞55 \frac {\sqrt{1+\frac 1 {\infty^2}}-\sqrt[3]{\frac 1 \infty+\frac 1 {\infty^3}}} {\sqrt[4]{1+\frac 1 {\infty^4}}- \sqrt[5]{\frac 1 \infty+\frac 1 {\infty^5}}} = \frac {\sqrt{1+\cancel{\frac 1 {\infty^2}}}-\sqrt[3]{\cancel{\frac 1 \infty}+\cancel{\frac 1 {\infty^3}}}} {\sqrt[4]{1+\cancel{\frac 1 {\infty^4}}}- \sqrt[5]{\cancel{\frac 1 \infty}+\cancel{\frac 1 {\infty^5}}}}

Granična vrednost je 1. 1 .

1+0−0+031+04−0+05=11=1 \frac {\sqrt{1+0}-\sqrt[3]{0+0}} {\sqrt[4]{1+0}- \sqrt[5]{0+0}} = \frac 1 1=1

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