Find the limit $\lim_{n\to\infty}\sqrt[n]{\dfrac{x^n}{2n+1}}=\text{?} \ \ \ \ :x>0$ 
Find the limit below:
$$\lim_{n\to\infty}\sqrt[n]{\dfrac{x^n}{2n+1}}=\text{?} \ \ \ \ :x>0$$

My Try :
$$a_n:=\sqrt[n]{\dfrac{x^n}{2n+1}}\\ \ln a_n=\dfrac{1}{n}\ln \left(\dfrac{x^n}{2n+1} \right)\\ \ln a_n=\dfrac{1}{n}\left(n\ln x-\ln(2n+1)\right) \\ \ln  a_n=\ln x-\ln(\dfrac{2n+1}{n})  $$
So we have :

$$\lim_{n\to\infty}\sqrt[n]{\dfrac{x^n}{2n+1}}=e^{\ln x-\ln 2}$$

Is it  right ?

you are almost right the only problem is $\frac{1}{n} \ln (2n+1)$ is NOT equal to $\ln(\frac{2n+1}{n})$. But $\frac{1}{n}\ln(2n+1)\to 0$ when $n\to \infty$. Then
$$
\ln(a_n)=\ln(x) +\frac{1}{n}\ln(2n+1)\to \ln(x)
$$ 
Hence $a_n$ tends to $x$.
