Question

$\underset{x\to 0}{lim}\frac{e^{2x}-1}{\sin 3x}=$

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$\frac{2}{3}$

Now,

$\underset{x\to 0}{lim}\frac{e^{2x}-1}{\sin 3x}$ $\left(\frac{0}{0}\right)$ form

Now applying L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{2e^{2x}}{3\cos 3x}$

$=\frac{2}{3}$.