How to evaluate a limit of the indeterminate form $(0/0)^0$ How to find the $\lim_{n \to \infty} \left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$? 
I know how to find it for the indeterminate form of $1^{\infty}$ by converting it into $0/0$ form, but this cannot be converted into any known indeterminate form: $(0/0)^0$ . Can we convert it into an integral and then try to solve at as infinity is involved?. May someone help? Also please don't use any theorems or formula for limits except perhaps L'Hospital rule which I know. If you do please provide its proof as well.
 A: First "trick:" convert to the exponential form.
$$
\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right)^{\frac{1}{n}}
=\exp\left({\frac{1}{n}\ln\left(\frac{\prod_{i=1}^{2n}(n+k)}{n^{2n}}\right)}\right)
$$
Now, let us focus on the exponent:
$$
\frac{1}{n}\ln\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right) = 
\frac{1}{n}\ln\left(\prod_{k=1}^{2n}\frac{n+k}{n}\right)
= \frac{1}{n}\ln\left(\prod_{k=1}^{2n}\left(1+\frac{k}{n}\right)\right)
= \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+\frac{k}{n}\right)
$$
At that point, it starts to really look like a Riemann sum, so let us massage it a little bit more:
$$
\frac{1}{n}\ln\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right) 
= \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+\frac{k}{n}\right)
= \frac{2}{2n}\sum_{k=1}^{2n}\ln\left(1+2\frac{k}{2n}\right)
= 2\cdot\frac{1}{2n}\sum_{k=1}^{2n}f\!\left(\frac{k}{2n}\right)
$$
for $f\colon [0,1]\to \mathbb{R}$ defined by $f(x)=\ln(1+2x)$.
We have$^{(\dagger)}$ $$\frac{1}{2n}\sum_{k=1}^{2n}f\!\left(\frac{k}{2n}\right)\xrightarrow[n\to\infty]{} \int_0^1 f = \frac{1}{2}(3\ln 3 -2)$$
and by continuity of the exponential your limit will be 
$$\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right)^{\frac{1}{n}}
\xrightarrow[n\to\infty]{} e^{2\int_0^1 f} = e^{3\ln 3 -2)} = \frac{27}{e^2}.$$

$(\dagger)$ Here, we use the following theorem, which essentially follows from the definition of Riemann integration:

Theorem. (Riemann sums converge to the integral.) Let $f\colon [a,b]\to\mathbb{R}$ be a continuous function. Then
  $$
\frac{1}{n}\sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right) \xrightarrow[n\to\infty]{} \int_a^b f
$$
  and
  $$
\frac{1}{n}\sum_{k=0}^{n-1} f\left(a+k\frac{b-a}{n}\right) \xrightarrow[n\to\infty]{} \int_a^b f.
$$

This is a particular case of a slightly more general theorem ($f$ only needs to be Riemann integrable, and here we took a regular subdivision of $[a,b]$ in intervals of the same length $\frac{b-a}{n}$ instead of an arbitrary subdivision.)
In our case, $a=0$ and $b=1$, and we take a subdivision with $m\stackrel{\rm def}{=} 2n$ points:
$$
\frac{1}{m}\sum_{k=1}^m f\left(\frac{k}{m}\right) \xrightarrow[m\to\infty]{} \int_0^1 f.
$$
A: Maybe you can use the exponential form.
Forgive me for what I'm going to write, I know it's bad formalism but it may be effective:
$$\left(\frac{0}{0}\right)^0 = e^{0\cdot \ln\left(\frac{0}{0}\right)}$$
Then you may use hospital for the limit inside the logarithm and proceed...

Answer:-
$$ \text{let } y = \left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$$
$$\implies\log_e y = {1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)$$
$$\implies e^{{1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)} = y$$

$$\text{let } z = \log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)$$
$$\implies z = \log_e (3n)!- \log_e n! - 2n\log_e n$$
$$\color{red}{\implies z = 3n\log_e (3n) - 3n - (n\log_e n - n) - 2n\log_e n}$$
$$\implies z = 3n\log_e (n) + 3n\log_e 3 - 2n - n\log_e n  - 2n\log_e n$$
$$\implies z =  n(3\log_e 3 - 2) $$

$$\text{substituting the value of z in y}$$
$$y= e^{{1\over n}z}$$
$$\implies y= e^{3\log_e 3 - 2}$$
$$\text{Finally finding the limit}$$
$$\implies \lim_{n \to \infty} y = \lim_{n \to \infty} e^{3\log_e 3 - 2} = e^{3\log_e 3 - 2} = {e^{3\log_e 3}\over e^2} = {27\over e^2}$$
$$\color{red}{RED} \leftarrow \text{Stirling approximation}$$
A: HINT:
Let $A=\left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$
$$\implies\ln A=\dfrac1n\sum_{r=1}^{2n}\ln\left(1+\dfrac rn\right)$$
Now like Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$, let $2n=m$
$$\implies\ln A=2\cdot\dfrac1m\sum_{r=1}^m\ln\left(1+2\cdot\dfrac rm\right)$$
$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\implies\ln A=2\int_0^1\ln(1+2x)\ dx$$
Now integrate by parts,
$$\int\ln(1+2x)\ dx=\ln(1+2x)\int dx-\int\left(\dfrac{d\{\ln(1+2x)\}}{dx}\cdot\int dx\right)dx$$
$$=x\ln(1+2x)-\int\dfrac x{1+2x}dx$$
Now for $\int\dfrac x{1+2x}dx,$ set $1+2x=y$ to ultimately find that
$$\ln A=2\left(\dfrac32\ln(1+2)-1\right)=3\ln 3-2=\ln\dfrac{3^3}{e^2}\text{ as }\ln(e)=1$$
