Question

$\underset{x\to 0}{lim}\frac{e^{ax}-e^{-2ax}}{\ln (1+x)}=$

• A. $a$
• B. $2a$
• C. $3a$
• D. $4a$

Answer 1

The correct option is C $3a$

Clearly form of the limit is $\frac{0}{0}$
Thus using L-Hospital's rule,
$\underset{x\to 0}{lim}\frac{e^{ax}-e^{-2ax}}{\ln (1+x)}=\underset{x\to 0}{lim}\frac{a{e}^{ax}+2a{e}^{-2ax}}{\frac{1}{1+x}}=a+2a=3a$