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