Problem
\( \lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x} \)
Solution
Step 1 ：\( \lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x} = \frac{1}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \)
Step 2 ：\( \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} = \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \cdot \frac{2}{\sin 2 x} \)
Step 3 ：\( \frac{1}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} = \frac{1}{5} \cdot 1 \)
