Question

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

• A. $e$
• B. $e^{1/2}$
• C. $e^{-2}$
• D. none of these

Answer 1

The correct option is D none of these

$\underset{x\to 0}{lim}{\left(\frac{1+5x^2}{1+3x^2}\right)}^{1/x^2}$
$=\underset{x\to 0}{lim}\frac{(1+5x^2{)}^{1/x^2}}{(1+3x^2{)}^{1/x^2}}$
Let $5x^2=y$, $3x^2=t$
$=\underset{y\to 0t\to 0}{lim}\frac{(1+y{)}^{5/y}}{(1+t{)}^{3/t}}$
$=\underset{y\to 0t\to 0}{lim}\frac{\left[\right(1+y{)}^{1/y}{]}^5}{\left[\right(1+t{)}^{1/t}{]}^3}$
$=\frac{e^5}{e^3}=e^2$ ($\because {lim}_{x\to 0}(1+x{)}^{1/x}=e$)