Limitless

Here is my proof by induction: Let $P_{n}$ be the proposition that $\large \displaystyle\lim_{x\rightarrow 1} \frac{(1-\sqrt{x})\cdot (1-\sqrt[3]{x})\cdots (1-\sqrt[n]{x})}{(1-x)^{n-1}} =\,0$ for positive integer n.

For base case n=1:

LHS= $\lim_{x \rightarrow 1}\frac{(1-\sqrt{x})}{(1-x)^{0}}=0$

Assume $P_{k}$ is true for some positive integer k.

Then, when $n=k+1$ :

LHS= $\large \displaystyle \lim_{x\rightarrow 1} \frac{(1-\sqrt{x})\cdot (1-\sqrt[3]{x})\cdots (1-\sqrt[k]{x})(1-\sqrt[k+1]{x})}{(1-x)^{k}} = \large \displaystyle\lim_{x\rightarrow 1} \frac{(1-\sqrt{x})\cdot (1-\sqrt[3]{x})\cdots (1-\sqrt[k]{x})}{(1-x)^{k-1}}\cdot \frac{1-\sqrt[k+1]{x}}{x^{k}}$

$\large \displaystyle = \lim_{x\rightarrow 1} \frac{(1-\sqrt{x})\cdot (1-\sqrt[3]{x})\cdots (1-\sqrt[k]{x})}{(1-x)^{k-1}}\cdot \lim_{x \rightarrow 1}\frac{1-\sqrt[k+1]{x}}{x^{k}}$

$\large \displaystyle =0$

Hence, by Mathematical Induction, since $P_{1}$ , $P_{k}$ , $P_{k+1}$ are true, the proof is complete.