<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] 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.
</s>