find $\lim_{x \to 1}\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)...\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{\left(n-1\right)}}$

Question

$$\lim_{x \to 1}\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)...\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{\left(n-1\right)}}$$

I've tried solving the limit by factoring, but it made the solution more difficult, also I cannot find any pattern,I was wondering if there exist a solution which does not use Taylor series.

Answer 1

Use $$\lim_{x\to1}\frac{1-x^{1/k}}{1-x}=\lim_{y\to1}\frac{1-y}{1-y^k} =\lim_{y\to1}\frac1{1+y+\cdots+y^{k-1}}=\frac1k$$ for $2\le k\le n$.

Answer 2

We have that

$$\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)...\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{\left(n-1\right)}}=\prod_{k=2}^n \frac{1-\sqrt[k]{x}}{1-x}$$

and by $f_k(x)=\sqrt[k]{x}$

$$\lim_{x \to 1}\frac{1-\sqrt[k]{x}}{1-x}=\lim_{x \to 1}\frac{\sqrt[k]{x}-1}{x-1}=f'_k(1)=\frac1k$$

Answer 3

Hint :

Consider $\lim f(x) = \exp (\lim \log f(x))$. Then divide your limit by $n-1$ parts and calculate it.