Question

The value of $\underset{x\to 0}{lim}\frac{x\ln (1+7x)}{1-\cos 3x}$ is

• A. $\frac{7}{5}$
• B. $\frac{14}{5}$
• C. $\frac{14}{9}$
• D. $\frac{5}{9}$

Answer 1

The correct option is C $\frac{14}{9}$

$L=\underset{x\to 0}{lim}\frac{\left(\frac{\ln (1+7x)}{x}\right)}{\left(\frac{1-\cos 3x}{x^2}\right)}\Rightarrow L=\underset{x\to 0}{lim}\frac{(7\cdot \frac{\ln (1+7x)}{7x})}{\left(\frac{\frac{9}{4}\cdot 2{\sin }^2\frac{3x}{2}}{{\left(\frac{3x}{2}\right)}^2}\right)}\therefore L=\frac{14}{9}$