Bài tập 28 trang 211 SGK Toán 11 NC

a)

\(\begin{array}{*{20}{l}}
{\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x}}{{\sin 5x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{\cos 2x.\sin 5x}}}\\
{ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}.\frac{1}{{\cos 2x.\frac{{\sin 5x}}{{5x}}}}.\frac{2}{5} = \frac{2}{5}}
\end{array}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{x\sin 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{2x\sin x\cos x}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{2x\cos x}} = \frac{1}{2}
\end{array}\)

c)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{1 + \sin x - \cos x}}{{1 - \sin x - \cos x}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\sin }^2}\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}}}
\end{array}\\
{ = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2} + \cos \frac{x}{2}}}{{\sin \frac{x}{2} - \cos \frac{x}{2}}} =  - 1}
\end{array}\)

-- Mod Toán 11 HỌC247