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.
 A: Hint: First show $(1+1/n +o(1/n))^n \to e.$
A: Use the fact that $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$. It is going to be quite lengthy so bear with me: The numerator is
$$
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n-
\left(1+\frac{1}{n}-\frac{1}{n^2}\right)^n = \frac{2}{n^2}\sum_{m=0}^{n-1}
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1-m}
\left(1+\frac{1}{n}-\frac{1}{n^2}\right)^m
$$
Write the above as $2/n^2 f(n)$.
The denominator has two parts
$$
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n-
\left(1+\frac{1}{n}-\frac{1}{n^2+1}\right)^n = \left(\frac{1}{n^2}+\frac{1}{n^2+1}\right)g(n)
$$
where
$$g(n) = \sum_{m=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1+m}\left(1+\frac{1}{n}-\frac{1}{n^2+1}\right)^m$$
and the second part
$$
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n-
\left(1+\frac{1}{n}-\frac{1}{n^2(n^2+1)}\right)^n = \frac{1}{n^2}\left(1+\frac{1}{n^2+1}\right)k(n)
$$
where
$$k(n)=\sum_{m=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1+m}\left(1+\frac{1}{n}-\frac{1}{n^2(n^2+1)}\right)^m$$
so in the end you have
$$
\lim_{n\to\infty}\frac{\frac{2}{n^2}f(n)}{\frac{2n^2+1}{n^2(n^2+1)}g(n)+\frac{n^2+2}{n^2(n^2+1)}h(n)}=
\lim_{n\to\infty}\frac{2(n^2+1)f(n)}{(2n^2+1)g(n)+(n^2+2)h(n)}
$$
Now note that as $n\to \infty$ then $f(n)$ becomes approximately
$$f(n) = \sum_{m=0}^{n-1}
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1-m}
\left(1+\frac{1}{n}-\frac{1}{n^2}\right)^m\approx n\left(1+\frac{1}{n}\right)^n\approx ne$$
where $\approx$ means to order of $o(1)$. Actually $g(n)\approx ne\approx h(n)$ too.
 So the limit reduces to
$$\to \lim_{n\to\infty}\frac{2(n^2+1)}{(2n^2+1)+(n^2+2)}=\frac{2}{3}$$
Hopefully I haven't made a mistake anywhere, but at least I think the method is clear.
A: 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 $2/3$.
A: Hint: (for the first version, before multiplying by 2)
$$
\left(1+\frac1n\right)^n\leq\left(1+\frac1n+\frac1{n^2}\right)^n\leq\left(1+\frac{1+\epsilon}n\right)^n
$$
for large $n$. Hence all partial limits are $e$ and the final one — 0.
A: 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}
A: $$
n\log\left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)=1+\frac xn-\frac1{2n}+O\left(\frac1{n^2}\right)
$$
Therefore,
$$
\left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)^n=e\left(1+\frac xn-\frac1{2n}\right)+O\left(\frac1{n^2}\right)
$$
Thus
$$
\begin{align}
&\frac{\overbrace{\left(1+\frac1n+\frac1{n^2}\right)^n}^{x=1}-\overbrace{\left(1+\frac1n-\frac1{n^2}\right)^n}^{x=-1}}{2\underbrace{\left(1+\frac1n+\frac1{n^2}\right)^n}_{x=1}-\underbrace{\left(1+\frac1n-\frac1{n^2+1}\right)^n}_{x=-1}-\underbrace{\left(1+\frac1n-\frac1{n^2(n^2+1)}\right)^n}_{x=0}}\\
&=\frac{e\left(1+\frac1n-\frac1{2n}\right)-e\left(1-\frac1n-\frac1{2n}\right)+O\left(\frac1{n^2}\right)}{2e\left(1+\frac1n-\frac1{2n}\right)-e\left(1-\frac1n-\frac1{2n}\right)-e\left(1-\frac1{2n}\right)+O\left(\frac1{n^2}\right)}\\[6pt]
&=\frac{\frac2n+O\left(\frac1{n^2}\right)}{\frac3n+O\left(\frac1{n^2}\right)}
\end{align}
$$
