_ n [ n^2 + 1 ] = - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \sin [\pi \sqrt {{n^2} + 1} ] = $

A
$\infty $
✓
$0$
C
Does not exist
D
None of these

✓

Answer

Correct option: B.

$0$

b
(b) Given limit $ = \mathop {\lim }\limits_{n \to \infty } \,\,\sin \left\{ {n\pi {{\left( {1 + \frac{1}{{{n^2}}}} \right)}^{1/2}}} \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } \,\,\sin \,\left\{ {n\pi \left( {1 + \frac{1}{{2{n^2}}} - \frac{1}{{8{n^4}}} + ...} \right)} \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } \,\,\sin \,\left\{ {n\pi \left( {1 + \frac{1}{{2n}} - \frac{1}{{8{n^3}}} + ...} \right)} \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } \,\,{( - 1)^n}\,\sin \pi \,\left( {\frac{1}{{2n}} - \frac{1}{{8{n^3}}} + ....} \right) = 0.$

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