Convergence and limit of $\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}$ as $n \to \infty$? I was calculating in WA some values of some sequences and constructed this  sequence:
$$a_n=\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}$$
It is immediate that expression in parentheses tends to $0$ since both terms tend to $1$, but the exponent also tends to $0$ so it may be that in the limit this behaves similarly to $x^x$ when $x$ tends to $0$ from the right side, but I am not sure.
How to prove existence of this limit and what is its value?
 A: Define $f(x) := (1/x)^x$. Then we have
$$
(n + 3)^\frac{1}{n + 3} - (n + 4)^\frac{1}{n + 4} = f\left(\frac{1}{n + 3}\right) - f\left(\frac{1}{n + 4}\right).
$$
By the mean value theorem
$$
f\left(\frac{1}{n + 3}\right) - f\left(\frac{1}{n + 4}\right) = f'(c_n) \left(\frac{1}{n + 3} - \frac{1}{n + 4}\right)
$$
for some $c_n \in (\frac{1}{n + 4}, \frac{1}{n + 3})$. So we have
$$
a_n = \left(\frac{f'(c_n)}{(n + 3)(n + 4)}\right)^{\frac{1}{n + 5}}.
$$
Now we can compute the derivative of $f$ to be
$$
f'(x) = \left(\frac{1}{x}\right)^x \left(\log\left(\frac{1}{x}\right) - 1\right).
$$
If we define
$$
b_n := \left(\frac{1}{(n + 3)(n + 4)}\right)^{\frac{1}{n + 5}},
$$
then we see that $\lim_{n \rightarrow \infty} b_n = 1$, so it remains to compute
$$
\lim_{n \rightarrow \infty} f'(c_n)^{\frac{1}{n +5}}
$$
and this should be straightforward.
A: First note that
\begin{align*}
a_n=\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}\leq(n+3)^{1/(n+3)}.
\end{align*}
Moreover, from Bernoulli's Inequality, we obtain
\begin{align*}
a_n&=\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}\\
&\geq\left(1-\frac{(n+4)^{1/(n+4)}}{(n+3)^{1/(n+3)}}\right)^{1/(n+5)} \\
&\geq 1-\frac{1}{(n+5)}\cdot\frac{(n+4)^{1/(n+4)}}{(n+3)^{1/(n+3)}}.
\end{align*}
Since both estimates go to 1 as $n\to\infty$, we get $a_n\to 1$.
