The value of the limit $\mathop {\lim }\limits_{x \to 0} {\left\{ {{1^{\frac{1}{{{{\sin }^2}x}}}} + {2^{\frac{1}{{{{\sin }^2}x}}}} + \dots + {n^{\frac{1}{{{{\sin }^2}x}}}}} \right\}^{{{\sin }^2}x}}$ is
- A$\infty $
- B$0$
- C$\frac{n(n + 1)}{2}$
- D$n$
Explore More
Similar Questions
Let $x_0$ be the real number such that $e^{x_0}+x_0=0$. For a given real number $\alpha$,define $g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)}$ for all real numbers $x$. Then which one of the following statements is True?
Let $f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3$,$x \in R$. Then the natural number $n$ for which $\lim_{x \rightarrow 1} \frac{x^{n} f(1) - f(x)}{x - 1} = 44$ is ...... .
DifficultJEE MAIN 2021
View SolutionEvaluate $\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n} - [x]}}{{[x]}},$ where $n \in N$ and $[x]$ denotes the greatest integer less than or equal to $x$.
$\mathop {\lim }\limits_{x \to a} \frac{{({x^{ - 1}} - {a^{ - 1}})}}{{x - a}} = $
Easy
View SolutionIf $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo