I tried solving the following problem many times, but each time I do I end up with the wrong answer. Is the order which I differentiated this expression in correct? Also, is the derivative of e (a constant) e or 0? I'm still very new to calculus, so I'd really appreciate a detailed explanation in regarsds to solving this problem. Thanks.

$\displaystyle \displaystyle 2(1-e^2){\lim_{n \to \infty}(\frac{\frac{e^{2/n}}{n}}{1 - e^{2/n}})$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(n(e^{2/n})' - (e^{2/n})n')}{n^2}}{1' - (\frac{2}{n}(e^{2/n - 1})e')}$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(n(\frac{2}{n}(e^{2/n - 1})e) - (e^{2/n})1)}{n^2}}{-(\frac{2}{n}(e^{2/n - 1})e)}$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(2(e^{2/n}) - (e^{2/n})}{n^2}}{-(\frac{2}{n}(e^{2/n})}$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}(\frac{(2(e^{2/n}) - (e^{2/n})}{n^2})(\frac{1}{-{2}{n}(e^{2/n})})$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{2(e^{2/n}) - (e^{2/n})}{n^2(-\frac{2}{n}(e^{2/n}))}$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{2(e^{2/n}) - (e^{2/n})}{(-2n(e^{2/n}))}$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{(e^{2/n})(2 - 1)}{-2n(e^{2/n})}$

$\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{1}{-2n}$

$\displaystyle = \displaystyle 2(1-e^2)(0)$

$\displaystyle = 0 $