Функције – граничне вредности функција 6

Текст задатака објашњених у видео лекцији.

Одредити граничне вредности:

пр.18)   $\mathop {\lim }\limits_{x \to \infty } {(1 + \frac{5}{x})^x}$

пр.19)   $\mathop {\lim }\limits_{x \to 0} {(1 + 2x)^{\frac{3}{{5x}}}}$

пр.20)   $\mathop {\lim }\limits_{x \to \infty } {(\frac{{1 - x}}{{7 - x}})^{\frac{2}{3}x}}$

пр.18)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } {(1 + \frac{5}{x})^x} = {\left( {1 + \frac{5}{\infty }} \right)^\infty } = {1^\infty } \hfill \\
\boxed{\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{x}} \right)}^x} = e} \hfill \\
\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{\frac{x}{5}}}} \right)^{\frac{x}{5} \cdot \frac{5}{x} \cdot x}} = {e^5} \hfill \\
\end{gathered} \]

пр.19)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} {(1 + 2x)^{\frac{3}{{5x}}}} = {\left( {1 + 0} \right)^{^{\frac{3}{{5 \cdot 0}}}}} = {1^\infty } \hfill \\
\boxed{\mathop {\lim }\limits_{x \to 0} {{\left( {1 + x} \right)}^{\frac{1}{x}}} = e} \hfill \\
\mathop {\lim }\limits_{x \to 0} {(1 + 2x)^{\frac{1}{{2x}} \cdot \frac{{2x}}{1} \cdot \frac{3}{{5x}}}} = {e^{\frac{6}{5}}} = \sqrt[5]{{{e^6}}} \hfill \\
\end{gathered} \]

пр.20)  

\[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{1 - x}}{{7 - x}}} \right)^{\frac{2}{3}x}} = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{{1 - x}}{{7 - x}} - 1} \right)^{\frac{2}{3}x}} = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{{1 - x - 8 + x}}{{7 - x}}} \right)^{\frac{2}{3}x}} = \hfill \\
= \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{{ - 6}}{{7 - x}}} \right)^{\frac{2}{3}x}} = \hfill \\
\boxed{\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{x}} \right)}^x} = e} \hfill \\
= \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{\frac{{7 - x}}{{ - 6}}}}} \right)^{\frac{{7 - x}}{{ - 6}} \cdot \frac{{ - 6}}{{7 - x}} \cdot }}^{\frac{2}{3}x} = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{{ - 4x}}{{7 - x}} \cdot \frac{x}{x}}} = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{ - 4x}}{x}}}{{\frac{{7 - x}}{x}}}}} = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{{ - 4}}{{\frac{7}{x} - 1}}}} = {e^{\frac{{ - 4}}{{ - 1}}}} = {e^4} \hfill \\
\end{gathered} \]