<s>[INST] 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.
 [/INST] Perhaps a  way to 'see the answer' is to use the MVT: for the numerator, write $f( x) = x^n$, $a = 1 + 1/n -1/n^2$, $b = 1 + 1/n + 1/n^2$. Then 
$$ f( b) - f(a) = f'(c)(b-a),$$
for some $c \in (a,b)$.
 Therefore the numerator is
$$ n\left(1 + 1/n + o(1/n^2)\right)^{n-1} 2/n^2.$$
Similarly, in the denominator, using the same trick twice, one gets
$$ n\left(1 +1/n + o(1/n^2)\right)^{n-1} ( 2/n^2 + 1/(n^2+1) + 1/n^2(n^2+1) ).$$
Taking the limit (of the ratio!), one gets Hamed's answer of $2/3$.
</s>