Question

$\displaystyle \sf\lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{n} \sqrt[k]{cos(kx)} }{ {x}^{2} } =$
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Answer 1

Answer:

Step-by-step explanation:

Let,

$u = \prod_{k=2}^{n} \sqrt[k]{coskx}\\or,\log u=\sum_{k=2}^{n} \frac{1}{k}\log{coskx}\\\text{On differentiating both sides,we get}\\\frac{1}{u}\cdot \frac{du}{dx} = \sum_{k=2}^{n} \frac{1}{k}\bigg(-\frac{kcos{kx}}{cos{kx}}\bigg)\\or, \frac{du}{dx} = \prod_{k=2}^{n} \sqrt[k]{coskx}\cdot \sum_{k=2}^{n} {-tan{kx}}~~~(\because u = \prod_{k=2}^{n} \sqrt[k]{coskx})$

Now,

$\displaystyle \sf\lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{n} \sqrt[k]{cos(kx)} }{ {x}^{2}}$

$=\displaystyle \sf\lim_{x \to 0 } \frac{ -\prod_{k=2}^{n} \sqrt[k]{coskx}\cdot \sum_{k=2}^{n} {(-tan{kx})}}{2x}~~~(\text{using L'Hospital's rule})\\\\= \displaystyle \sf\lim_{x \to 0 } \frac{1}{2}\cdot \[\frac{(tan2x+tan3x+\ldots+tannx)}{x}\\= \frac{1}{2}\cdot (2+3+4+\ldots +n) = \frac{n^2+n-2}{4}$

Answer 2

Answer:

Step-by-step explanation:

Let,

$u = \prod_{k=2}^{n} \sqrt[k]{coskx}\\or,\log u=\sum_{k=2}^{n} \frac{1}{k}\log{coskx}\\\text{On differentiating both sides,we get}\\\frac{1}{u}\cdot \frac{du}{dx} = \sum_{k=2}^{n} \frac{1}{k}\bigg(-\frac{kcos{kx}}{cos{kx}}\bigg)\\or, \frac{du}{dx} = \prod_{k=2}^{n} \sqrt[k]{coskx}\cdot \sum_{k=2}^{n} {-tan{kx}}~~~(\because u = \prod_{k=2}^{n} \sqrt[k]{coskx})$

Now,

$\displaystyle \sf\lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{n} \sqrt[k]{cos(kx)} }{ {x}^{2}}$

$=\displaystyle \sf\lim_{x \to 0 } \frac{ -\prod_{k=2}^{n} \sqrt[k]{coskx}\cdot \sum_{k=2}^{n} {(-tan{kx})}}{2x}~~~(\text{using L'Hospital's rule})\\\\= \displaystyle \sf\lim_{x \to 0 } \frac{1}{2}\cdot \[\frac{(tan2x+tan3x+\ldots+tannx)}{x}\\= \frac{1}{2}\cdot (2+3+4+\ldots +n) = \frac{n^2+n-2}{4}$