Вычислить предел от тригонометрической функции
Задача. Вычислить предел заданной функции.
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{tgx - tg2}}{{\sin \ln (x - 1)}} = \left( {\frac{0}{0}} \right) = \left| \begin{array}{l}2 - x = y\\y \to 0\end{array} \right| = \mathop {\lim }\limits_{y \to 0} \frac{{tg(2 - y) - tg2}}{{\sin \ln (1 - y)}} = \\ = \mathop {\lim }\limits_{y \to 0} \frac{{tg(2 - y) - tg2}}{{\ln (1 - y)}} = = \mathop {\lim }\limits_{y \to 0} \frac{{\frac{{ - 1}}{{{{\cos }^2}\left( {2 - y} \right)}}}}{{\frac{{ - 1}}{{1 - y}}}} = \\ = \mathop {\lim }\limits_{y \to 0} \frac{{1 - y}}{{{{\cos }^2}(2 - y)}} = \frac{1}{{{{\cos }^2}2}} = 1 + t{g^2}2.\end{array}\]
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{tgx - tg2}}{{\sin \ln (x - 1)}} = \left( {\frac{0}{0}} \right) = \left| \begin{array}{l}2 - x = y\\y \to 0\end{array} \right| = \mathop {\lim }\limits_{y \to 0} \frac{{tg(2 - y) - tg2}}{{\sin \ln (1 - y)}} = \\ = \mathop {\lim }\limits_{y \to 0} \frac{{tg(2 - y) - tg2}}{{\ln (1 - y)}} = = \mathop {\lim }\limits_{y \to 0} \frac{{\frac{{ - 1}}{{{{\cos }^2}\left( {2 - y} \right)}}}}{{\frac{{ - 1}}{{1 - y}}}} = \\ = \mathop {\lim }\limits_{y \to 0} \frac{{1 - y}}{{{{\cos }^2}(2 - y)}} = \frac{1}{{{{\cos }^2}2}} = 1 + t{g^2}2.\end{array}\]
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