How to compute this gross limit. 
How do I compute this limit?
  $$
  \lim_{n \to \infty} 
  \frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n - 
        \left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
  }{
       2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n - 
        \left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n - 
        \left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n
  }
$$

I think I got the correct limit by using fast converging limits to $e$.
In particular I used truncated Taylor series for the sqrt and 4th root.
Or squares and bisquares.
Example
$(1+1/2n)^{2n}$
 Becomes 
$(1 + 1/n + 1/4n^2)^n.$
In combination with l'hopital it gives me the answer.
But I guess that is not a very good (fast) method.

You might get some easy simplifications by extending the fraction with the factor $(1-\frac1n)^n$ so that all terms transform to the form
$$
\left(1+\frac{c_n}{n^2}\right)^n=1+\frac{c_n}{n}+O(n^{-2}).
$$
to obtain
\begin{align}
&\ \lim_{n \to \infty} 
  \frac{\left(1  - \frac{1}{n^3}\right)^n - 
        \left(1  - \frac{2}{n^2} + \frac{1}{n^3}\right)^n
  }{
       2 \left(1 - \frac{1}{n^3}\right)^n - 
        \left(1 - \frac{2n^2+1-n}{n^2 (n^2 + 1)}\right)^n - 
        \left(1 - \frac{n^2+2}{n^2 (n^2+1)} + \frac{1}{n^3 (n^2 +1)}\right)^n
  }
\\ \\
&=\lim_{n \to \infty} 
  \frac{\left(1 + O(n^{-2})\right) - 
       \left(1-\frac2n +  O(n^{-2}) \right)
  }{
       2 \left(1 +  O(n^{-2})\right) - 
        \left(1 - \frac{2}{n} + O(n^{-2})\right) - 
        \left(1 - \frac{1}{n} + O(n^{-2})\right)
  }
\\ \\
&=\frac 23
\end{align}
