Question

$\underset{x\to 0}{lim}{\left(\frac{1+2x}{1+3x}\right)}^{\frac{1}{x^2}}.e^{\frac{1}{x}}=$

• A. $e^{\frac{5}{2}}$
• B. $e^2$
• C. 2
• D. 1

Answer 1

The correct option is A $e^{\frac{5}{2}}$

$\underset{x\to 0}{lim}{e}^{\frac{1}{x^2}\left(log\right(1+2x)-log(1+3x\left)\right)+\frac{1}{x}}{e}^{\underset{x\to 0}{lim}\frac{\left(log\right(1+2x)-log(1+3x\left)\right)+x}{x^2}}$
Using the expansion we get -
$=e^{\frac{5}{2}}$