Question

The value of $\underset{x\to 0}{lim}{\left(\begin{array}{c}\frac{1+5x^2}{1+3x^2}\end{array}\right)}^{x^{\frac{1}{2}}}$ is

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

Answer 1

The correct option is A $e^2$

$\underset{x\to 0}{lim}{\left(\begin{array}{c}\frac{1+5x^2}{1+3x^2}\end{array}\right)}^{x^{\frac{1}{2}}}=e^{\underset{x\to 0}{lim}\frac{1}{2}\left(\begin{array}{c}\frac{1+5x^2}{1+3x^2}-1\end{array}\right)}$
$=e^{\underset{x\to 0}{lim}\frac{2x^2}{(1+3x^2)}}=e^2$