Question

Solve${lim}_{x\leftarrow 0}\frac{\ln \left(\tan (\frac{\pi }{4}+ax)\right)}{\sin bx},b\ne 0$ is equal to

• A. $\frac{a}{b}$
• B. $\frac{2a}{b}$
• C. $\frac{a}{2b}$
• D. $\frac{b}{a}$

Answer 1

Answer: (B)

$\frac{2a}{b}$

if we put x=0 we get undetermined form so apply L hospitol rule

now we get $\frac{\frac{a}{tan(\frac{\pi }{4}+ax)}.se{c}^2(\frac{\pi }{4}+ax)}{bcosbx}$

put x=0

$\frac{2a}{b}$