Question

$\underset{x\to 0}{lim}\frac{\log (\sin 7x+\cos 7x)}{\sin 3x}$ equals:

• A. $\frac{7}{3}$
• B. $\frac{14}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{3}\log 7$

Answer 1

The correct option is A $\frac{7}{3}$

$\underset{x\to 0}{lim}\frac{\log (\sin 7x+\cos 7x)}{\sin 3x}=\frac{1}{2}\underset{x\to 0}{lim}\frac{\log (1+\sin 14x)}{\sin 3x}$

$=\frac{1}{2}\underset{x\to 0}{lim}\{\frac{\log (1+\sin 14x)}{\sin 14x}.\frac{\sin 14x}{14x}.\frac{14x}{3x}.\frac{3x}{\sin 3x}\}=\frac{1}{2}.\frac{14}{3}=\frac{7}{3}$