Question

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

• A. $\frac{2}{3}$
• B. 6
• C. $\frac{3}{2}$
• D. $\frac{1}{6}$

Answer 1

The correct option is A $\frac{2}{3}$

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

Using L-Hospital's rule, we get

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