Q130 / 200 · Manipal (MET) 2014 · Mathematics

$\lim _{x \rightarrow 0}\left(1^{\operatorname{cosec}^2 x}+2^{\operatorname{cosec}^2 x}+\ldots+n^{\operatorname{cosec}^2 x}\right)^{\sin ^2 x}$ is equal to

  • A
    1
  • B
    $\frac{1}{n}$
  • Correct answer
    C
    $n$
  • D
    0

Solution

Let $y=\operatorname{cosec}^2 x$, $\begin{aligned} & \text { Required limit }=\lim _{y \rightarrow \infty}\left(y^y+2^y+\ldots n^y\right)^{1 / y} \\ & =\lim _{y \rightarrow \infty}\left(n^y\right)^{1 / y} \quad(0 / \infty \text { form }) \\ & \qquad\left[\left(\frac{1}{n}\right)^y+\left(\frac{2}{n}\right)^y+\ldots+\left(\frac{n-1}{n}\right)^y+1\right]^{1 / y} \\ & =\lim _{y \rightarrow \infty} n\left[\left(\frac{1}{n}\right)^y+\left(\frac{2}{n}\right)^y+\ldots+\left(\frac{n-1}{n}\right)^y+1\right]^{1 / y} \\ & =n \cdot 1^0=n \cdot 1=n \end{aligned}$

Hence, the correct answer is option: 'C'.

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