MCQ
$\lim_\limits{\text{n} \rightarrow \infty}​\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$​ is equal to:
  • A
    $0$
  • B
    $2$
  • $4$
  • D
    $ \infty$

Answer

Correct option: C.
$4$
$=\lim_\limits{\text{n} \rightarrow \infty}​\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$
$ = \displaystyle \lim_{\text{n}\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{\text{n}} \right)^2}{\left(1+\Large \frac{2}{\text{n}} \right)\left(1+\Large \frac{3}{\text{n}} - \Large \frac{1}{\text{n}^2} \right)} }$
$=\text{n}$
$ = \displaystyle \frac{(2+0)^2}{(1+0)(1+0+0)}$

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