This is the question
If $$\lim_{n\to\infty}\left(\frac{\left(1+\frac1n\right)^n}{\left(1-\frac1n\right)^n}-e^2\right)n^2=\frac{ae^2}{b}$$ for $a, b\in\Bbb{N}$, then find the minimum value of $a+b$.
My approach - I first plugged $1/n=h$ and changed the overall question accordingly. After that I wrote the first complex term as $e^{Ln}(x=e^{\ln x}$ property) ,then multiplied $e^2$ and divided to that term and removed $e^2$ common from the whole term. Now I am stuck. Kindly help or suggest an alternative approach. Thank you.
Replacing $n$ with $\frac{1}{x}$ and dividing through by $e^2$, your problem becomes $$ \mathop {\lim }\limits_{x \to 0} \frac{{\exp \left( {\frac{1}{x}\log \left( {\frac{{1 + x}}{{1 - x}}} \right) - 2} \right) - 1}}{{x^2 }}=\frac{a}{b}. $$ But \begin{align*} \mathop {\lim }\limits_{x \to 0} \frac{{\exp \left( {\frac{1}{x}\log \left( {\frac{{1 + x}}{{1 - x}}} \right) - 2} \right) - 1}}{{x^2 }} &= \mathop {\lim }\limits_{x \to 0} \frac{{\exp \left( {2\left( {\frac{{\tanh ^{ - 1} x}}{x} - 1} \right)} \right) - 1}}{{x^2 }} \\ & = \mathop {\lim }\limits_{x \to 0} \frac{{\exp \left( {2\left( {\frac{{\tanh ^{ - 1} x}}{x} - 1} \right)} \right) - 1}}{{2\left( {\frac{{\tanh ^{ - 1} x}}{x} - 1} \right)}}\frac{2}{{x^2 }}\left( {\frac{{\tanh ^{ - 1} x}}{x} - 1} \right). \end{align*} I assume we know that $$ \mathop {\lim }\limits_{x \to 0} \frac{{\tanh ^{ - 1} x}}{x} = 1,\quad \mathop {\lim }\limits_{w \to 0} \frac{{e^w - 1}}{w} = 1, $$ so the limit simplifies to $$ \mathop {\lim }\limits_{x \to 0} \frac{2}{{x^2 }}\left( {\frac{{\tanh ^{ - 1} x}}{x} - 1} \right). $$ This limit can be computed by using the standard Maclaurin series of the function $\tanh ^{ - 1} x$ or by L'Hôpital's rule (if you do not want to use series at all).