Question

$\underset{x\to 0}{\lim }{\left(\frac{1+5x^2}{1+3x^2}\right)}^{\frac{1}{x^2}}=$

• A. $e^{3x}$
• B. $e^2$
• C. $\frac{1}{e}$
• D. $\frac{5}{3}$

Answer 1

Answer: (B)

Let $f\left(x\right)={\left(\frac{1+5x^2}{1+3x^2}\right)}^{\frac{1}{x^2}}$
$\therefore {\lim }_{x\to 0}f\left(x\right)={\lim }_{x\to 0}{\left(\frac{1+5x^2}{1+3x^2}\right)}^{\frac{1}{x^2}}$
$=e^{{\lim }_{x\to 0}{\left(\frac{1+5x^2}{1+3x^2}\right)}^{\frac{1}{x^2}}}$
$=e^{{\lim }_{x\to 0}{\left(\frac{1+5x^2}{1+3x^2}\right)}^{\frac{1}{x^2}}}=e^2$