Question

limits question

100 points {please answer}

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Answer 1

Answer:

Step-by-step explanation:

lim x tends to 0 {1^1/sin^2(x)+..........+n^1/sin2^x}

lim x tends to 0{1^(sin^-1(x))^2=..........+n^(sin^-(x))^2}

substitute x=0 we get

1^(sin^-(0))^2+......+n^(sin^-1(0))^2}

we know sin inverse 0 is 0

and 0 ka square is 0

so we get expression as 1^0+2^0+.......+n^0(upto n)

1+1+1+......+1(n numbers)

n

Answer 2

Answer:

    Option (3), n(n + 1)/2 is correct

Step-by-step explanation:

  We have to find the value of

$\displaystyle \mathsf{\lim_{x\to 0}\{1^{1/Sin^{2}x}+2^{1/Sin^{2}x}+...+n^{1/Sin^{2}x}\}^{Sin^{2}x}}$

$\displaystyle \mathsf{=\lim_{x\to 0}\big(\sum_{n=1}^{n}n^{1/Sin^{2}x}\big)^{Sin^{2}x}}$

Let, $\displaystyle \mathsf{y=\big(\sum_{n=1}^{n}n^{1/Sin^{2}x}\big)^{Sin^{2}x}}$

Then $\displaystyle \mathsf{logy=Sin^{2}x\:.\:log\big(\sum_{n=1}^{n}n^{1/Sin^{2}x}\big)}$

Taking limit on both sides as $\displaystyle \mathsf{x\to 0}$ , we get

$\displaystyle \mathsf{\lim_{x\to 0}\:logy=\lim_{x\to 0}Sin^{2}x\:.\:log\big(\sum_{n=1}^{n}n^{1/Sin^{2}x}\big)}$

$\displaystyle \mathsf{=\lim_{x\to 0}\frac{log\big(\sum_{n=1}^{n}n^{Cosec^{2}x}\big)}{Cosec^{2}x}\:[\frac{\infty}{\infty}\:form]}$

$\displaystyle \mathsf{=\lim_{x\to 0}\frac{\frac{\big(\displaystyle \mathsf{\sum_{n=1}^{n}n^{Cosec^{2}x}.logn.(-2\:Cosec^{3}x\:Cotx)\big)}}{\big(\displaystyle \mathsf{\sum_{n=1}^{n}n^{Cosec^{2}x}\big)}}}{-2\:Cosec^{3}x\:Cotx}}$

$\displaystyle \mathsf{=\sum_{n=1}^{n}logn}$

$\displaystyle \mathsf{= log(1 + 2 + 3 + ... + n)}$

$\implies \displaystyle \mathsf{\lim_{x\to 0}logy=log\:\lim_{x\to 0}y= log(1 + 2 + ... + n)}$

This gives $\displaystyle \mathsf{\lim_{x\to 0}y=1+2+...+n}$

$\displaystyle\mathsf{=\frac{n(n+1)}{2}}$ ,

  which is the required limiting value.