Question

The value of $\underset{x\to 0}{lim}\frac{1}{x}\log \left(\frac{1+ax}{1-bx}\right)$ is

• A. $0$
• B. $\frac{a}{b}$
• C. $a+b$
• D. $e^{\frac{a}{b}}$

Answer 1

The correct option is C $a+b$

$\underset{x\to 0}{lim}\frac{1}{x}\log \left(\frac{1+ax}{1-bx}\right)$

$\underset{x\to 0}{lim}\frac{\log (1+ax)-\log (1+bx)}{x}$ (Using log properties)

$\underset{x\to 0}{lim}a\left(\frac{\log (1+ax)}{ax}\right)-(-b)\left(\frac{\log 1+(-b)x}{(-b)x}\right)$

$\underset{x\to 0}{lim}a\left(\frac{\log (1+ax)}{ax}\right)-\underset{x\to 0}{lt}(-b)\left(\frac{\log 1+(-b)x}{(-b)x}\right)$

$=a-(-b)$

$=a+b$