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Функције - граничне вредности функција 8

Гранична вредност функције. Лимеси тригонометријских, логаритамских и експоненцијалних функција, сложенији примери.

Задаци

Текст задатака објашњених у видео лекцији.

Одредити граничне вредности:

пр.26)   $\mathop {\lim }\limits_{x \to 0} \frac{{tg x - \sin x}}{{{{\sin }^3}x}}$

пр.27)   $\mathop {\lim }\limits_{x \to 0} \frac{{\ln (a + x) - lna}}{x}$

пр.28)   $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{2x}}$

пр.29)   $\mathop {\lim }\limits_{x \to 0} \frac{{{e^{ax}} - {e^{bx}}}}{x}$

пр.26)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{tg} x - \sin x}}{{{{\sin }^3}x}} = \frac{{\operatorname{t} g0 - \sin 0}}{{{{\sin }^3}0}} = \frac{{0 - 0}}{0} = \frac{0}{0} \hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{{\operatorname{tg} x - \sin x}}{{{{\sin }^3}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin x}}{{\cos x}} - \sin x}}{{{{\sin }^3}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin x - \sin x\cos x}}{{\cos x}}}}{{\frac{{{{\sin }^3}x}}{1}}} = \hfill \\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin x\cos x}}{{{{\sin }^3}x\cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x\left( {1 - \cos x} \right)}}{{{{\sin }^3}x\cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos x} \right)}}{{\left( {1 - {{\cos }^2}x} \right)\cos x}} = \hfill \\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)\cos x}} = \frac{1}{2} \hfill \\
\end{gathered} \]

пр.27)

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{\ln (a + x) - lna}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\ln (\frac{{a + x}}{a})}}{{\frac{x}{a}a}} = \frac{1}{a} \hfill \\
\boxed{\mathop {\lim }\limits_{x \to 0} \frac{{\ln (x + 1)}}{x} = 1} \hfill \\
\end{gathered} \]

пр.28)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \frac{1}{{{e^x}}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{{\left( {{e^x}} \right)}^2} - 1}}{{{e^x}}}}}{{\frac{{2x}}{1}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{2x{e^x}}} = \frac{1}{{{e^0}}} = \frac{1}{1} = 1 \hfill \\
\boxed{\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1} \hfill \\
\end{gathered} \]

пр.29)   

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{{e^{ax}} - {e^{bx}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{ax}} - 1 + 1 - {e^{bx}}}}{x} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{ax}} - 1}}{x} + \frac{{1 - {e^{bx}}}}{x}} \right) = \hfill \\
= \mathop {\lim }\limits_{x \to 0} \frac{{{e^{ax}} - 1}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{1 - {e^{bx}}}}{x} = a + \mathop {\lim }\limits_{x \to 0} \frac{{1 - {e^{bx}}}}{{bx}}b = a + \left( { - b} \right) = a - b \hfill \\
\boxed{\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1} \hfill \\
\end{gathered} \]