I need some help with the following expression: $$ \lim_{n \rightarrow \infty}(5+n)^{2n-1}((n+1)^{{1}/{n}}-n^{{1}/{n}}).$$
What I have done so far is the following:
\begin{align*} (5+n)^{2n-1} ((n+1)^{\frac{1}{n}}-n^{\frac{1}{n}})&= (1+\frac{5}{n})^{2n-1} \cdot n^{2n-1} \cdot ((1+\frac{1}{n})^\frac{1}{n}-1)^n \cdot n\\ &= (1+\frac{5}{n})^{2n-1} \cdot n^{2n} \cdot ((1+\frac{1}{n})^\frac{1}{n}-1)^n \end{align*}
Now I would like to use that $ \lim_{n \rightarrow \infty} (1+\frac{x}{n})^n = e^x $ but i dont know exactly how to use it in this case.
We can use it for an asymptotic analysis to guess what the limit is, indeed
$$(5+n)^{2n-1} = n^{2n-1}\left(1+\frac 5{n}\right)^{2n-1}=n^{2n-1}\left[\left(1+\frac 5{n}\right)^{n}\right]^\frac{2n-1}{n}\sim e^{10}n^{2n-1}$$
and since
$$(n+1)^{{1}/{n}}=e^{\frac{\log (n+1)}{n}} \sim 1+\frac{\log(n+1)}n$$
$$n^{{1}/{n}}=e^{\frac{\log n}{n}} \sim 1+\frac{\log n}n$$
we have
$$((n+1)^{{1}/{n}}-n^{{1}/{n}})\sim \frac{\log(n+1)}n-\frac{\log n}n=\frac1n \log \frac{n+1}{n}=\frac1{n^2}\log\left(1+\frac1n\right)^n$$
therefore
$$(5+n)^{2n-1}((n+1)^{{1}/{n}}-n^{{1}/{n}}) \sim \frac{e^{10}n^{2n-1}\log\left(1+\frac1n\right)^n}{n^2} \to \infty$$