Looking for the limit of a sequence How to find the limit of the sequence 
$$
\left(\frac{2}{2p+1}\right)^{1/p}
$$
as $p \to \infty$?
Is it just $1$?
 A: Hint: you can write your sequence (I write $n\in \mathbb N$ instead of $p$) as
$$\lim _{n\to \infty} (2^{\frac{1}{n}} (2n+1)^{-\frac{1}{n}})=\lim _{n\to \infty}(2^{\frac{1}{n}})\cdot \lim _{n\to \infty}  (2n+1)^{-\frac{1}{n}}$$
$$=\lim _{n\to \infty} (2^{\frac{1}{n}})\cdot \lim _{n\to \infty}\left(e^{-\frac{1}{n}\ln \left(2n+1\right)}\right)=1\cdot 1=1$$
Remember that: $$\lim _{n\to \infty}-\frac{\ln (2n+1)}{n}=0\tag{1}$$
because $-\frac{\ln (2n+1)}{n}=-\frac{2\ln \sqrt{2n+1}}{n}<2\sqrt{\frac{2n+1}{n^2}}$
and being $\lim _{n\to \infty} 2\sqrt{\frac{2n+1}{n^2}}=0 \implies (1)$
A: Using L'Hospital,
$$\lim_{p\to\infty}\frac{\log(p+\frac12)}p=\lim_{p\to\infty}\frac1{p+\frac12}=0$$ and the initial limit is $1$.
A: $\dfrac{1}{(3p)^{1/p}}< (\dfrac{2}{2p+1})^{1/p}<$
$(\dfrac{2p+1}{2p+1})^{1/p}=1$
Take the limit.
Recall : $\lim_{p \rightarrow \infty} p^{1/p}=1.$
A: We have that
$$\left(\frac{2}{2p+1}\right)^{1/p}=\frac{2^{1/p}}{(2p+1)^{1/p}}$$
and since $2^{1/p}\to 2^0=1$ we need to consider
$$\lim_{p\to \infty}(2p+1)^{1/p} =1$$
indeed
$$\left[(2p+1)^{\frac1{2p+1}}\right]^{\frac{2p+1}p}\to 1^2 =1$$
since we need to recall the foundamental result as $x\to \infty$
$$x^\frac1x=e^{\frac{\log x}x} \to e^0=1$$
indeed by $x=e^y$ with $y\to \infty$
$$\frac{\log x}x=\frac y{e^y}\to 0$$
