Question

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

• A. $a+b$
• B. $a-b$
• C. $-(a+b)$
• D. $ab$

Answer 1

Answer: (B)

$a-b$

$L=\underset{x\to 0}{lim}\frac{\log (1+ax)-\log (1+bx)}{x}$
Applying L' Hospital rule
$L=\underset{x\to 0}{lim}\frac{\frac{a}{1+ax}-\frac{b}{1+bx}}{1}$
$=a-b$