Question

\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\] 

Answer 1

\[\lim_{x \to 0} \left[ \frac{\sin \left( 2x \right)}{e^x - 1} \right]\] 

Dividing the numerator and the denominator by x: 

\[= \lim_{x \to 0} \left[ \frac{\sin 2x}{x \times \frac{e^x - 1}{x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin 2x}{2x} \times \frac{2}{\left( \frac{e^x - 1}{x} \right)} \right]\]
\[ = 1 \times \frac{2}{1}\]
\[ = 2\]