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Zadatak 0847

Test: MKsIVBGI_073

 

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

 

Rešenje:


 

Najpre izračunajmo granične vrednosti brojioca i imenioca posebno:

$$\lim \limits_{n \to +\infty }\sqrt{n+\sqrt{n+\sqrt{n}}}-\sqrt{n}\cdot \frac{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}$$

$$=\lim \limits_{n \to +\infty }\frac{n+\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}$$

$$=\lim \limits_{n \to +\infty }\frac{\sqrt{n+\sqrt{n}}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}$$

$$=\lim \limits_{n \to +\infty }\frac{1}{\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}{\sqrt{n+\sqrt{n}}}}$$

$$=\lim \limits_{n \to +\infty }\frac{1}{\sqrt{\frac{n}{n+\sqrt{n}}+\frac{\sqrt{n+\sqrt{n}}}{n+\sqrt{n}}}+\sqrt{\frac{1}{\frac{n+\sqrt{n}}{n}}}}$$

$$=\lim \limits_{n \to +\infty }\frac{1}{\sqrt{\frac{1}{1+\frac{1}{\sqrt{n}}}+\frac{1}{\sqrt{n+\sqrt{n}}}}+\sqrt{\frac{1}{1+\frac{1}{\sqrt{n}}}}}$$

$$=\frac{1}{2}$$

Za limes imenioca imamo:

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

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

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

$$=\frac{1}{\infty }=0$$

Napokon: $$\lim \limits_{n \to +\infty }\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}-\sqrt{n}}{\sqrt[3]{8n^3+1}-2n}=0$$


 

 

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