MCQ
The value of $\mathop {\lim }\limits_{n \to \infty } \frac{{1 + 2 + 3 + ....n}}{{{n^2} + 100}}$ is equal
  • A
    $\infty $
  • $\frac{1}{2}$
  • C
    $2$
  • D
    $0$

Answer

Correct option: B.
$\frac{1}{2}$
b
(b) We have, $\mathop {\lim }\limits_{n \to \infty } \frac{{1 + 2 + 3 + ..... + n}}{{{n^2} + 100}}$

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{n(n + 1)}}{{2({n^2} + 100)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}\left( {1 + \frac{1}{n}} \right)}}{{2{n^2}\left( {1 + \frac{{100}}{{{n^2}}}} \right)}} = \frac{1}{2}$.

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