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Функције - граничне вредности функција 3

Гранична вредност функције. Неодређеност бесконачно кроз бесконачно, сложенији примери.

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Одредити граничне вредности:

пр.9)   $\mathop {\lim }\limits_{x \to \infty } \frac{{(x - 1)(x - 2)(x - 3)(x - 4)}}{{{{(2x - 5)}^4}}}$

пр.10)   $\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 1}}{{\sqrt {2{x^4} + 2} }}$

пр.11)   $\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt[4]{{x + {x^4}}} - \sqrt[5]{{3 + {x^2}}}}}{{\sqrt[3]{x} - x - \sqrt[4]{{1 - x}}}}$

 

пр.9)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{(x - 1)(x - 2)(x - 3)(x - 4)}}{{{{(2x - 5)}^4}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{(x - 1)(x - 2)(x - 3)(x - 4)}}{{{{(2x - 5)}^4}}} \cdot \frac{{{x^4}}}{{{x^4}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{(x - 1)(x - 2)(x - 3)(x - 4)}}{{{x^4}}}}}{{\frac{{{{(2x - 5)}^4}}}{{{x^4}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{x - 1}}{{{x^4}}} \cdot \frac{{x - 2}}{{{x^4}}} \cdot \frac{{x - 3}}{{{x^4}}} \cdot \frac{{x - 4}}{{{x^4}}}}}{{{{\left( {\frac{{2x - 5}}{x}} \right)}^4}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {1 - \frac{1}{x}} \right)\left( {1 - \frac{2}{x}} \right)\left( {1 - \frac{3}{x}} \right)\left( {1 - \frac{4}{x}} \right)}}{{{{(2x - 5)}^4}}} = \frac{{1 \cdot 1 \cdot 1 \cdot 1}}{{{2^4}}} = \frac{1}{{16}} \hfill \\
\end{gathered} \]

пр.10)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 1}}{{\sqrt {2{x^4} + 2} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 1}}{{\sqrt {2{x^4} + 2} }} \cdot \frac{{{x^2}}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{2{x^2} - 1}}{{{x^2}}}}}{{\frac{{\sqrt {2{x^4} + 2} }}{{\sqrt {{x^4}} }}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{2{x^2} - 1}}{{{x^2}}}}}{{\frac{{\sqrt {2{x^4} + 2} }}{{\sqrt {{x^4}} }}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2 - \frac{1}{{{x^2}}}}}{{\sqrt {\frac{{2{x^4} + 2}}{{{x^4}}}} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{2 - \frac{1}{{{x^2}}}}}{{\sqrt {2 + \frac{2}{{{x^4}}}} }} = \frac{{2 - 0}}{{\sqrt {2 + 0} }} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \hfill \\
\end{gathered} \]

пр.11)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt[4]{{x + {x^4}}} - \sqrt[5]{{3 + {x^2}}}}}{{\sqrt[3]{x} - x - \sqrt[4]{{1 - x}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt[4]{{x + {x^4}}} - \sqrt[5]{{3 + {x^2}}}}}{{\sqrt[3]{x} - x - \sqrt[4]{{1 - x}}}} \cdot \frac{x}{x} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{\sqrt[4]{{x + {x^4}}}}}{x} - \frac{{\sqrt[5]{{3 + {x^2}}}}}{x}}}{{\sqrt[3]{{\frac{x}{{{x^3}}}}} - 1 - \sqrt[4]{{\frac{{1 - x}}{{{x^4}}}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt[4]{{\frac{x}{{{x^4}}} + 1}} - \sqrt[5]{{\frac{3}{{{x^5}}} + \frac{{{x^2}}}{{{x^5}}}}}}}{{\sqrt[3]{{\frac{x}{{{x^3}}}}} - 1 - \sqrt[4]{{\frac{{1 - x}}{{{x^4}}}}}}} = \hfill \\
= \frac{{\sqrt[4]{{0 + 1}} - \sqrt[5]{0}}}{{\sqrt[3]{0} - 1 - \sqrt[4]{0}}} = - 1 \hfill \\
\end{gathered} \]