Zadatak 0848

Test: MKsIVBGI_073

 

Izračunaj graničnu vrednost niza: $$\lim \limits_{n \to +\infty }\frac{\sqrt[4]{n^4+2n}-n}{\sqrt[3]{8n^3+1}-2n}$$

 

Rešenje:


$$...=\lim \limits_{n \to +\infty }\frac{\sqrt[4]{n^4+2n}-n}{\sqrt[3]{8n^3+1}-2n}\cdot \frac{\sqrt[4]{n^4+2n}+n}{\sqrt[4]{n^4+2n}+n}\cdot \frac{\sqrt[3]{(8n^3+1)^2}+2n\sqrt[3]{8n^3+1}+4n^2}{\sqrt[3]{(8n^3+1)^2}+2n\sqrt[3]{8n^3+1}+4n^2}$$

$$=\lim \limits_{n \to +\infty }\frac{\sqrt{n^4+2n}-n^2}{8n^3+1-8n^3}\cdot \frac{\sqrt[3]{(8n^3+1)^2}+2n\sqrt[3]{8n^3+1}+4n^2}{\sqrt[4]{n^4+2n}+n}\cdot \frac{\sqrt{n^4+2n}+n^2}{\sqrt{n^4+2n}+n^2}$$

$$=\lim \limits_{n \to +\infty }\frac{\left ( n^4+2n-n^4 \right )\left ( \sqrt[3]{(8n^3+1)^2}+2n\sqrt[3]{8n^3+1}+4n^2 \right )}{(\sqrt[4]{n^4+2n}+n)\left ( \sqrt{n^4+2n}+n^2 \right )}$$

$$=2\lim \limits_{n \to +\infty }\frac{n\cdot n^2\left ( \sqrt[3]{\left ( 8+\frac{1}{n^3} \right )^2} +2\sqrt[3]{8+\frac{1}{n^3}}+4\right )}{n\left ( \sqrt[4]{1+\frac{2}{n^3}}+1 \right )\cdot n^2\left ( \sqrt{1+\frac{2}{n^3}}+1 \right )}$$

$$=2\cdot \frac{4+4+4}{2\cdot 2}=6$$