Question

$\underset{x\to 0}{lim}(1^{{\text{cosec}}^{2}x}+2^{cose{c}^{2}x}+...+n^{cose{c}^{2}x}{)}^{{\sin }^{2}x}$ is equal to

• A. $1$
• B. $\frac{1}{4}$
• C. $n$
• D. $0$

Answer 1

The correct option is C $n$

Let $y={\text{cosec}}^{2}x$,
Required limit $\underset{y\to \infty }{lim}(1^y+2^y+....n^y{)}^{1/y}(0/\infty$ for)
$=\underset{y\to \infty }{lim}(n^y{)}^{1/y}$
$[{\left(\frac{1}{n}\right)}^y+{\left(\frac{2}{n}\right)}^y+....+{\left(\frac{n-1}{n}\right)}^y+1]$
$=\underset{y\to \infty }{lim}n[{\left(\frac{1}{n}\right)}^y+{\left(\frac{2}{n}\right)}^y+....+{\left(\frac{n-1}{n}\right)}^y+1]$
$=n\cdot 1^0$

$=n\cdot 1=n$