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Zadatak 0845

Test: MKsIVBGI_073

 

Izračunaj graničnu vrednost niza: $$\lim \limits_{n \to +\infty }\frac{\sqrt[3]{2n^2+1}-\sqrt[3]{2n^2-1}}{\sqrt{4n^2+11}-\sqrt{4n^2-11}}$$

 

Rešenje:


$$...=\lim \limits_{n \to +\infty }\frac{\sqrt[3]{2n^2+1}-\sqrt[3]{2n^2-1}}{\sqrt{4n^2+11}-\sqrt{4n^2-11}}\cdot \frac{\sqrt[3]{(2n^2+1)^2}+\sqrt[3]{4n^4-1}+\sqrt[3]{(2n^2-1)^2}}{\sqrt[3]{(2n^2+1)^2}+\sqrt[3]{4n^4-1}+\sqrt[3]{(2n^2-1)^2}}\cdot \frac{\sqrt{4n^2+11}+\sqrt{4n^2-11}}{\sqrt{4n^2+11}+\sqrt{4n^2-11}}$$

$$=\lim \limits_{n \to +\infty }\frac{(2n^2+1-2n^2+1)\left ( \sqrt{4n^2+11}+\sqrt{4n^2-11} \right )}{(4n^2+11-4n^2+11)\left ( \sqrt[3]{(2n^2+1)^2}+\sqrt[3]{4n^4-1}+\sqrt[3]{(2n^2-1)^2} \right )}$$

$$=\frac{1}{11}\lim \limits_{n \to +\infty }\frac{n\left ( \sqrt{4+\frac{11}{n^2}}+\sqrt{4-\frac{11}{n^4}} \right )}{\sqrt[3]{n^4}\left ( \sqrt[3]{(2+\frac{1}{n^2})^2}+\sqrt[3]{4-\frac{1}{n^4}}+\sqrt[3]{(2-\frac{1}{n^4})^2} \right )}$$

$$=\frac{1}{11}\cdot 0=0$$


 

 

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