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.

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}
