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Zadatak 0846

Test: MKsIVBGI_073

 

Izračunaj graničnu vrednost niza: $$\lim \limits_{n \to +\infty }\frac{\sqrt[4]{n^4+4}-n}{2n-\sqrt[4]{16n^4+1}}$$

 

Rešenje:


 

$$...=\lim \limits_{n \to +\infty }\frac{\sqrt[4]{n^4+4}-n}{2n-\sqrt[4]{16n^4+1}}\cdot \frac{\sqrt[4]{n^4+4}+n}{\sqrt[4]{n^4+4}+n}\cdot \frac{2n+\sqrt[4]{16n^4+1}}{2n+\sqrt[4]{16n^4+1}}$$

$$=\lim \limits_{n \to +\infty }\frac{\sqrt{n^4+4}-n^2}{4n^2-\sqrt{16n^4+1}}\cdot \frac{2n+\sqrt[4]{16n^4+1}}{\sqrt[4]{n^4+4}+n}\cdot \frac{\sqrt{n^4+4}+n^2}{\sqrt{n^4+4}+n^2}\cdot \frac{4n^2+\sqrt{16n^4+1}}{4n^2+\sqrt{16n^4+1}}$$

$$=\lim \limits_{n \to +\infty }\frac{n^4+4-n^4}{16n^4-16n^4-1}\cdot \frac{2n+\sqrt[4]{16n^4+1}}{\sqrt[4]{n^4+4}+n}\cdot \frac{4n^2+\sqrt{16n^4+1}}{\sqrt{n^4+4}+n^2}$$

$$=-4\lim \limits_{n \to +\infty }\frac{2n\left ( 1+\sqrt[4]{1+\frac{1}{16n^4}} \right )}{n\left ( \sqrt[4]{1+\frac{4}{n^4}}+1 \right )}\cdot \frac{4n^2\left (  1+\sqrt{1+\frac{1}{16n^4}}\right )}{n^2\left ( \sqrt{1+\frac{4}{n^4}}+1 \right )}$$

$$=-4\cdot 8=-32$$


 

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