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 $\frac{1}{2}$ is

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

Answer 1

Answer: (B)

$4e$

$y\left(n\right)=e^{x+x^2+...+x^n}=e^{\frac{x(1-x^n)}{1-x}}$

So $\frac{dy\left(n\right)}{dx}=e^{\frac{x(1-x^n)}{1-x}}\times \frac{d}{dx}\left(\frac{x(1-x^n)}{1-x}\right)$

$\underset{n\to \infty }{lim}\frac{dy\left(n\right)}{dx}=\underset{n\to \infty }{lim}{e}^{\frac{x}{1-x}-\frac{x^{n+1}}{1-x}}\frac{d}{dx}\left[\underset{n\to \infty }{lim}\frac{x}{1-x}-\frac{x^{n+1}}{1-x}\right]$

$=e^{\frac{x}{dx}}\times \frac{d}{dx}\left(\frac{x}{dx}\right)=e^{\frac{x}{1-x}}\frac{d}{dx}(-1+\frac{1}{1-x})$

$=e^{\frac{x}{1-x}}.\frac{d}{dx}\frac{1}{{(1-x)}^2}$

$\because$ for $x\in (0,1);x^n\to 0$

$\therefore \underset{n\to \infty }{lim}\frac{x^{n+1}}{1-x}=0$

$\Rightarrow \underset{n\to \infty }{lim}{\left(\frac{dy\left(n\right)}{dx}\right)}_{x=\frac{1}{2}}=4e$