How to evaluate $\lim\limits_{x\to1}\frac{(1-x^{1/2})(1-x^{1/3})\cdots(1-x^{1/n})} {(1-x)^{n-1}}$ I used substitution 
$t=1-x, x=1-t, t\rightarrow 0$
After the substitution:
$$\lim\limits_{t\to0}\frac{(1-(1-t)^{1/2})(1-(1-t)^{1/3})\cdots(1-(1-t)^{1/n})} {t^{n-1}}$$
How to use rationalization to get:
$$\lim\limits_{t\to0}\frac{t^{n-1}}{n! t^{n-1}}$$
Thanks for replies.
 A: The limit can be easily evaluated if we use the standard formula $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ for $a > 0$ and rational $n$. Putting $a = 1$ and $n = 1/k$ where $k$ is a positive integer we can see that $$\lim_{x \to 1}\frac{1 - x^{1/k}}{1 - x} = \lim_{x \to 1}\frac{x^{1/k} - 1}{x - 1} = \frac{1}{k}\tag{2}$$ Now our desired limit can be calculated as
\begin{align}
L &= \lim_{x \to 1}\frac{(1 - x^{1/2})(1 - x^{1/3})\cdots (1 - x^{1/n})}{(1 - x)^{n - 1}}\notag\\
&= \lim_{x \to 1}\frac{1 - x^{1/2}}{1 - x}\cdot \frac{1 - x^{1/3}}{1 - x}\cdots \frac{1 - x^{1/n}}{1 - x}\notag\\
&= \frac{1}{2}\cdot\frac{1}{3}\cdots\frac{1}{n} = \frac{1}{n!}
\end{align}
A: Hint: use
$$ 1-x^{1/k}=\frac{1-x}{\sum_{i=0}^{k-1}x^{\frac{i}{k}}}.$$
Then
$$ \lim\limits_{x\to1}\frac{(1-x^{1/2})(1-x^{1/3})...(1-x^{1/n})} {(1-x)^{n-1}}=\lim\limits_{x\to1}\frac{1}{\Pi_{i=2}^k\sum_{i=0}^{k-1}x^{\frac{i}{k}}}$$
and you will get the answer.
A: Let us first transform $\ln f(x)$ as follows
\begin{eqnarray*}
\ln f(x) &=&\ln \left( (1-x^{1/2})(1-x^{1/3})\cdots (1-x^{1/n})\right)
-(n-1)\ln (1-x) \\
&=&\sum_{k=2}^{n}\ln (1-x^{1/k})-\sum_{k=2}^{n}\ln (1-x) \\
&=&\sum_{k=2}^{n}\left( \ln (1-x^{1/k})-\ln (1-x)\right)  \\
&=&\sum_{k=2}^{n}\ln \left( \frac{1-x^{1/k}}{1-x}\right)  \\
&=&\sum_{k=2}^{n}\ln \left( \frac{1-\left( x^{1/k}\right) }{1-x}\times \frac{%
(1+\left( x^{1/k}\right) +\left( x^{1/k}\right) ^{2}+\cdots +\left(
x^{1/k}\right) ^{(k-1)})}{(1+\left( x^{1/k}\right) +\left( x^{1/k}\right)
^{2}+\cdots +\left( x^{1/k}\right) ^{(k-1)})}\right)  \\
&=&\sum_{k=2}^{n}\ln \left( \frac{1-(x^{1/k})^{k}}{\left( 1-x\right)
(1+\left( x^{1/k}\right) +\left( x^{1/k}\right) ^{2}+\cdots +\left(
x^{1/k}\right) ^{(k-1)})}\right)  \\
&=&\sum_{k=2}^{n}\ln \left( \frac{1}{1+\left( x^{1/k}\right) +\left(
x^{1/k}\right) ^{2}+\cdots +\left( x^{1/k}\right) ^{(k-1)}}\right)  \\
&=&-\sum_{k=2}^{n}\ln \left( 1+\left( x^{1/k}\right) +\left( x^{1/k}\right)
^{2}+\cdots +\left( x^{1/k}\right) ^{(k-1)}\right) .
\end{eqnarray*}
Il follows that 
\begin{eqnarray*}
\lim_{x\rightarrow 1}\ln f(x) &=&-\lim_{x\rightarrow 1}\sum_{k=2}^{n}\ln
\left( 1+\left( x^{1/k}\right) +\left( x^{1/k}\right) ^{2}+\cdots +\left(
x^{1/k}\right) ^{(k-1)}\right)  \\
&=&-\sum_{k=2}^{n}\lim_{x\rightarrow 1}\ln \left( 1+\left( x^{1/k}\right)
+\left( x^{1/k}\right) ^{2}+\cdots +\left( x^{1/k}\right) ^{(k-1)}\right) ,\
finite\ sum\  \\
&=&-\sum_{k=2}^{n}\ln \left( 1+\left( \lim_{x\rightarrow 1}x^{1/k}\right)
+\left( \lim_{x\rightarrow 1}x^{1/k}\right) ^{2}+\cdots +\left(
\lim_{x\rightarrow 1}x^{1/k}\right) ^{(k-1)}\right) ,\ continuity\ of\ \ln \
and\ powers\ of\ x \\
&=&-\sum_{k=2}^{n}\ln \left( 1+\left( 1\right) +\left( 1\right) ^{2}+\cdots
+\left( 1\right) ^{(k-1)}\right)  \\
&=&-\sum_{k=2}^{n}\ln \left( k\right)  \\
&=&-\ln (n!) \\
&=&\ln \left( \frac{1}{n!}\right) .
\end{eqnarray*}
Now, 
\begin{eqnarray*}
\lim_{x\rightarrow 1}f(x) &=&\lim_{x\rightarrow 1}\exp (\ln f(x)) \\
&=&\exp (\lim_{x\rightarrow 1}\ln f(x)),\ \ \ continuity\ of\ \exp  \\
&=&\exp \ln \left( \frac{1}{n!}\right)  \\
&=&\frac{1}{n!}.
\end{eqnarray*}
