Find $ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $ without using L'Hopital's rule.

Question

Let $ n $ be a positive integer greater than $ 1 $.

Find : $$ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $$

Without L'Hopital's rule or series expansion.

Here is What I did to solve the problem.

\begin{aligned}\displaystyle\lim_{x\to 0}{\displaystyle\frac{1-\prod\limits_{k=2}^{2n+1}{\sqrt[k]{1+\left(-1\right)^{k}x}}}{x}}&=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}-\prod\limits_{i=2}^{k}{\sqrt[i]{1+\left(-1\right)^{i}x}}}{x}}}\\&=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{1-\sqrt[k]{1+\left(-1\right)^{k}x}}{x}\displaystyle\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}}}\\ &=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\left(-1\right)^{k+1}}{\sum\limits_{j=0}^{k-1}{\sqrt[k]{1+\left(-1\right)^{k}x}^{j}}}\displaystyle\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}}}\\ &=\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\left(-1\right)^{k+1}}{k}} \\ &=\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k+1}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k}}\\ &=-1+\displaystyle\sum_{k=0}^{n}{\displaystyle\frac{1}{2k+1}}+\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{k}}\\ &=-1+\displaystyle\sum_{k=1}^{2n+1}{\displaystyle\frac{1}{k}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{k}}\\ \displaystyle\lim_{x\to 0}{\displaystyle\frac{1-\prod\limits_{k=2}^{2n+1}{\sqrt[k]{1+\left(-1\right)^{k}x}}}{x}}&=H_{2n+1}-H_{n}-1 \end{aligned}

What's your approach to solve the problem ?

Answer 1

By definition of the derivative, the desired limit is equal to $-f'(0)$ where $$f(x) = \sqrt{1+x} \sqrt[3]{1-x} \sqrt[4]{1+x} \cdots \sqrt[2n+1]{1-x}.$$ We can now use logarithmic differentiation to find $$f'(x) = f(x) \sum_{k=2}^{2n+1} \frac{(-1)^k}{k} (1 + (-1)^k x)^{-1}.$$ Plugging in $x = 0$ and $f(0) = 1$ will give the desired result.

Answer 2

By the binomial formula $$(1+(-1)^kx)^{1/k}=1+\frac{(-1)^k}kx+O(x^2),$$ so that $$\begin{align}\prod_{k=2}^{2n+1}(1+(-1)^kx)^{1/k} &=\prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}kx+o(x)\right)\\ &=\prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}kx\right)+o(x)\\ &=1+x\sum_{k=2}^{2n+1}\frac{(-1)^k}k+o(x) \end{align}$$ so that the desired limit is $$-\sum_{k=2}^{2n+1}\frac{(-1)^k}k$$

This differs from your answer. Have I made a mistake?

Answer 3

Note that the numerator is of the form $$1-abc\dots$$ where each of $a, b, c, \dots$ tends to $1$. We can rewrite the above as $$1-a+a(1-bc\dots)$$ and effectively the limit is converted into a finite sum of limits as $$\sum_{k=2}^{2n+1}\lim_{x\to 0}\frac{1-\sqrt[k]{1+(-1)^kx}}{x}$$ Clearly expression under limit above can be written as $$\frac{1-t^{1/k}}{1-t}\cdot\frac{1-t} {x}$$ where $t=1+(-1)^kx$ and the above tends to $$\frac{(-1)^{k+1}}{k}$$ and the desired limit equals $$-\sum_{k=2}^{2n+1}(-1)^k\cdot\frac{1}{k}$$

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This technique has been used here and here.