How do I solve the following limit?
$$\lim _{x\to \:1} \frac{\left(1-x^{1/2}\right)\left(1-x^{1/3}\right)\cdots \left(1-x^{1/n}\right)}{(1-x)^{n-1}}$$
The solution should be 0, but the source isn't reliable.
Thank you.
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Related : http://math.stackexchange.com/questions/889127/tough-limit-to-evaluate – lab bhattacharjee Aug 12 '14 at 15:59
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$$\cdots=\lim_{x\to1 }\prod_{k=2}^n\frac{x^{\frac{1}{k}}-1}{(x-1)}\underset{(1)}{=}\prod_{k=2}^n\lim_{x\to1 }\frac{x^{\frac{1}{k}}-1}{(x-1)}\underset{(2)}{=}\prod_{k=2}^n \frac{1}{k}=\frac{1}{n!}$$
Justifications:
$(1):$ Because the product is finite.
$(2):$ Because $$\lim_{x\to1 }\frac{x^{\frac{1}{k}}-1}{x-1}=\frac{d x^{\frac{1}{k}}}{dx}\Big|_{x=1}=\frac{1}{k}1^{\frac{1}{k}-1}=\frac{1}{k}$$
idm
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The given equation is same as $$\lim_{x\to1}\prod_{i=2}^n {1-x^{1/i}\over1-x}.$$ By using $$\lim_{a\to1}{1-a\over1-a^m}=\lim_{a\to1}{1\over1+a+\dots+a^{m-1}}={1\over m},$$ we get $${1\over n!}.$$ Does the answer really $0$? Or am I wrong? Don't you have $(n\to\infty)$ in addition?
Jaehyeon Seo
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@EulCan But it is in $\prod_{i=2}^n$, so we multiply it $n-1$ times. – Jaehyeon Seo Aug 12 '14 at 14:13
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Gotcha, thanks for explaining. I figured I'd missed something like that :) – Jam Aug 12 '14 at 14:18