MCQ
$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\Sigma {n^2}}}{{{n^3}}}} \right] = $
- A$ - \frac{1}{6}$
- B$\frac{1}{6}$
- ✓$\frac{1}{3}$
- D$ - \frac{1}{3}$
Note : Students should remember that
$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum n}}{{{n^2}}} = \frac{1}{2},\,\,\,\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^2}}}{{{n^3}}} = \frac{1}{3}$
and $\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^3}}}{{{n^4}}} = \frac{1}{4}.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$A:$ $^{\prime}$ the sum is even $^{\prime}$.
$B:$ $^{\prime}$the sum is a multiple of $3$$^{\prime}$
$C:$ $^{\prime}$the sum is less than $4 $$^{\prime}$
$D:$ $^{\prime}$the sum is greater than $11$$^{\prime}$.
Which pairs of these events are mutually exclusive ?