Функције – граничне вредности функција 7

Текст задатака објашњених у видео лекцији.

Одредити граничне вредности:

пр.21)   $\mathop {\lim }\limits_{x \to \infty } {(\frac{{2{x^3} + 2{x^2} + 1}}{{2{x^3} + {x^2} + 2x + 1}})^{3x}}$

пр.22)   $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{x}$

пр.23)   $\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin 3x}}$

пр.24)   $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}}{{\sin 4x}}$

пр.25)   $\mathop {\lim }\limits_{x \to 0 } \frac{{tg 6x}}{{tg 2x}}$

пр.21)   

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

пр.22)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{x} = \mathop {\lim }\limits_{x \to 0} 4\frac{{\sin 4x}}{{4x}} = 4\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{4x}} = 4 \cdot 1 = 4 \hfill \\
\boxed{\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \hfill \\
\end{gathered} \]

пр.23)   \[\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin 3x}} = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin 3x}}{x}} \right)^{ - 1}} = \mathop {\lim }\limits_{x \to 0} {\left( {3\frac{{\sin 3x}}{{3x}}} \right)^{ - 1}} = {\left( {3 \cdot 1} \right)^{ - 1}} = \frac{1}{3}\]

пр.24)   \[\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}}{{\sin 4x}} = \mathop {\lim }\limits_{x \to 0} \frac{{5x \cdot \frac{{\sin 5x}}{{5x}}}}{{4x \cdot \frac{{\sin 4x}}{{4x}}}} = \frac{5}{4}\]

пр.25)  

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{tg 6x}}{{tg 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin 6x}}{{\cos 6x}}}}{{\frac{{\sin 2x}}{{\cos 2x}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin 6x \cdot \cos 2x}}{{\cos 6x \cdot \sin 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin 6x}}{{\sin 2x}} = \hfill \\
= \mathop {\lim }\limits_{x \to 0} \frac{{6x \cdot \frac{{\sin 6x}}{{6x}}}}{{2x \cdot \frac{{\sin 2x}}{{2x}}}} = \frac{6}{2} = 3 \hfill \\
\end{gathered} \]