Knowing the result of a given limit, calculate the following after. What is its result?

Question

So I know for sure that by resolving the following limit I will have that result.

$$ \lim_{x\to 1}\frac{\sqrt[m]{x}-1}{x-1}=\frac{1}{m}$$

And now I have this: $$ l=\lim_{x\to 1}\frac{(1-x)(1-\sqrt[3]{x})(1-\sqrt[5]x)...(1-\sqrt[n]{x})}{(1-x)^{n-1}}.$$ How can I resolve the last limit using the upper relation?

Answer 1

Your limit is equal to

$$ \lim_{x\to1} \frac{1-\sqrt{x}}{1-x} \cdot \lim_{x\to1} \frac{1-\sqrt[3]{x}}{1-x}\cdot\cdots\cdot\lim_{x\to1}\frac{1-\sqrt[n]{x}}{1-x} \cdot \lim_{x\to 1} \frac1{(1-x)^{n-1-\left\lceil\frac{n}2\right\rceil}} = 1\cdot\frac13 \cdot \frac15\cdots\frac1n \cdot +\infty = +\infty$$

Answer 2

By the product rule, $$l=\prod_{k=2}^n\lim_{x\to 1}\left(\frac{\sqrt[k]x-1}{x-1}\right) =\prod_{k=2}^n\cdots.$$