Question

$\underset{x\to 0}{lim}\frac{\log (1+3x^2)}{x(e^{5x}-1)}=$

• A. $\frac{3}{5}$
• B. $\frac{5}{3}$
• C. $\frac{-3}{5}$
• D. $\frac{-5}{3}$
• E. 1

Answer 1

The correct option is A $\frac{3}{5}$

$\underset{x\to 0}{lim}\frac{\log (1+3x^2)}{x(e^{5x}-1)}$
$=\underset{x\to 0}{lim}\frac{\log (1+3x^2)}{3x^2}\times \frac{5x}{e^{5x}-1}\times \frac{3}{5}$
$=\frac{3}{5}\underset{x\to 0}{lim}\frac{\log (1+3x^2)}{3x^2}\times \underset{x\to 0}{lim}\frac{5x}{e^{5x}-1}$
$=\frac{3}{5}\times 1\times 1=\frac{3}{5}$
$(\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1\text{and}\underset{x\to 0}{lim}\frac{e^x-1}{x}=1)$