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Zadatak 0040

Test: MPsIVKVI_008

 

Odredi granične vrednosti: a) \(\lim_{x \to \infty }\left (\frac{x^2+5x+4}{x^2+8x+7}  \right )^x\);  b)  \(\lim_{x \to \infty }\left ( \frac{3x^2+x}{x^3-8}-\frac{2}{x-2} \right )\)

 

Rešenje:

 

a) \(...=\lim_{x \to \infty }\left (\frac{(x+1)(x+4}{(x+1)(x+7)}  \right )^x\) \(\lim_{x \to \infty }(\frac{x+4}{x+7})^x\)

 

\(=\lim_{x \to \infty }\left (\frac{x+4}{x+7}  \right )^x=\lim_{x \to \infty }\left (\frac{x+7-3}{x+7}  \right )^x\)

 

\(=\lim_{x \to \infty }\left (1+\frac{-3}{x+7}  \right ))^x=\lim_{x \to \infty }\left (\left (1+\frac{1}{\frac{x+7}{-3}}  \right )^{\frac{x+7}{-3}}  \right )^{-3}\cdot \left ( 1+\frac{-3}{x+7} \right )^{-7}\)

 

 

\(=e^{-3}\).

b)

\(\lim_{x \to \infty }\left ( \frac{3x^2+x}{(x-2)(x^2+2x+4)}-\frac{2}{x-2} \right )\) \(=\lim_{x \to \infty }\frac{ ( 3x^2+x )-2(x^2+2x+4)}{x^3-8}\)

 

\(=\lim_{x \to \infty }\frac{x^2-3x-8}{x^3-8}=\lim_{x \to \infty }\frac{\frac{1}{x}-\frac{3}{x^2}-\frac{8}{x^3}}{1-\frac{8}{x^3}}=0\).

 


 

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