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Zadatak 0840

Test: MKsIVBGI_073

 

Izračunaj graničnu vrednost niza: $$\lim \limits_{n \to +\infty }\frac{\sqrt{5n^2+3}-\sqrt{5n^2+1}}{\sqrt{3n^2+5}-\sqrt{3n^2+1}}$$

 

Rešenje:


$$...=\lim \limits_{n \to +\infty }\frac{\sqrt{5n^2+3}-\sqrt{5n^2+1}}{\sqrt{3n^2+5}-\sqrt{3n^2+1}}\cdot \frac{\sqrt{5n^2+3}+\sqrt{5n^2+1}}{\sqrt{5n^2+3}+\sqrt{5n^2+1}}\cdot \frac{\sqrt{3n^2+5}+\sqrt{3n^2+1}}{\sqrt{3n^2+5}+\sqrt{3n^2+1}}$$

$$=\lim \limits_{n \to +\infty }\frac{(5n^2+3-5n^2-1)\left ( \sqrt{3n^2+5}+\sqrt{3n^2+1} \right )}{(3n^2+5-3n^2-1)\left ( \sqrt{5n^2+3}+\sqrt{5n^2+1} \right )}$$

$$=\frac{1}{2}\lim \limits_{n \to +\infty }\frac{\sqrt{3+\frac{5}{n^2}}+\sqrt{3+\frac{1}{n^2}}}{\sqrt{5+\frac{3}{n^2}}+\sqrt{5+\frac{1}{n^2}}}$$

$$=\frac{1}{2}\cdot \frac{2\sqrt{3}}{2\sqrt{5}}$$

$$=\frac{1}{2}\cdot \frac{\sqrt{15}}{5}=\frac{\sqrt{15}}{10}$$


 

 

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