Question

If $y\left(n\right)=e^x{e}^{x^2}...e^{x^n},0<x<1$. Then $\underset{n\to \infty }{lim}\frac{dy\left(n\right)}{dx}$ at $x=\frac{1}{2}$ is

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

Answer 1

The correct option is B $4e$

$y\left(n\right)=e^{x+x^2+...+x^n}=e^{\frac{x(1-x^n)}{1-x}}$
So $\frac{dy\left(n\right)}{dx}=e^{\frac{x(1-x^n)}{1-x}}\times \frac{d}{dx}\left(\frac{x(1-x^n)}{1-x}\right)$
$\underset{n\to \infty }{lim}\frac{dy\left(n\right)}{dx}=\underset{n}{lim}{e}^{\frac{x}{1-x}-\frac{x^{n+1}}{1-x}}\frac{d}{dx}[\underset{n\to \infty }{lim}\frac{x}{1-x}-\frac{x^{n+1}}{1-x}]$
$=e^{\frac{x}{1-x}}\frac{d}{dx}\left(\frac{x}{1-x}\right)$
$=e^{\frac{x}{1-x}}\frac{1}{{(1-x)}^2}$
$\underset{n\to \infty }{lim}\frac{dy\left(n\right)}{dx}{|}_{x=\frac{1}{2}}=4e$