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  •   Find <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{1-\sqrt{1+x} \sqrt[3]{1-x} \cdot \cdots \sqrt[2 n+1]{1-x}}{x}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%

  Find
\mathrm{\lim _{x \rightarrow 0} \frac{1-\sqrt{1+x} \sqrt[3]{1-x} \cdot \cdots \sqrt[2 n+1]{1-x}}{x}}

Option: 1

0


Option: 2

-1


Option: 3

2


Option: 4

1


Answers (1)

best_answer

By definition of the derivative, the desired limit is equal to \mathrm{-f^{\prime}(0)} where

\mathrm{f(x)=\sqrt{1+x} \sqrt[3]{1-x} \sqrt[4]{1+x} \cdots \sqrt[2 n+1]{1-x} }.

We can now use logarithmic differentiation to find
\mathrm{f^{\prime}(x)=f(x) \sum_{k=2}^{2 n+1} \frac{(-1)^{k}}{k}\left(1+(-1)^{k} x\right)^{-1} }.

Plugging in \mathrm{x=0} and \mathrm{f(0)=1} will give the desired result.

Posted by

seema garhwal

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