Функције – граничне вредности функција 2

Текст задатака објашњених у видео лекцији.

Одредити граничне вредности:

пр.6)   $\mathop {\lim }\limits_{x \to \infty } \frac{{7{x^3} - 2x + 3}}{{3{x^3} + 2}}$

пр.7)   $\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - x + 1}}{{5x + 3}}$

пр.8)   $\mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} - 1}}{{{x^2}(7x+3)}}$

пр.6)  

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{7{x^3} - 2x + 3}}{{3{x^3} + 2}} \cdot \frac{{{x^3}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{7{x^3} - 2x + 3}}{{{x^3}}}}}{{\frac{{3{x^3} + 2}}{{{x^3}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{7{x^3}}}{{{x^3}}} - \frac{{2x}}{{{x^3}}} + \frac{3}{{{x^3}}}}}{{\frac{{3{x^3}}}{{{x^3}}} + \frac{2}{{{x^3}}}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{7 - \frac{2}{{{x^2}}} + \frac{3}{{{x^3}}}}}{{3 + \frac{2}{{{x^3}}}}} = \frac{{7 - 0 - 0}}{{3 + 0}} = \frac{7}{3} \hfill \\
\end{gathered} \]

пр.7)

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

пр.8)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} - 1}}{{{x^2}(7x + 3)}} \cdot \frac{{{x^3}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{5{x^2} - 1}}{{{x^3}}}}}{{\frac{{{x^2}(7x + 3)}}{{{x^3}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{5{x^2}}}{{{x^3}}} - \frac{1}{{{x^3}}}}}{{\frac{{7{x^3}}}{{{x^3}}} + \frac{{3{x^2}}}{{{x^3}}}}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{x} - \frac{1}{{{x^3}}}}}{{7 + \frac{3}{x}}} = \frac{{0 - 0}}{{7 + 0}} = \frac{0}{7} = 0 \hfill \\
\end{gathered} \]