A) \[0\]
B) \[1\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[-\frac{1}{\sqrt{2}}\]
Correct Answer: D
Solution :
| We have, |
| \[\underset{n\to \infty }{\mathop{\lim }}\,{{(-1)}^{n-1}}\sin \left( \pi \sqrt{{{n}^{2}}+\frac{n}{2}+1} \right)\] |
| \[=\underset{n\to \infty }{\mathop{\lim }}\,\sin \left( n\pi -\pi \sqrt{{{n}^{2}}+\frac{n}{2}+1} \right)\] |
| \[[\therefore sinx={{\left( -1 \right)}^{n-1}}\sin \left( n\pi -x \right)]\] |
| \[=\underset{n\to \infty }{\mathop{\lim }}\,\sin \left( \pi \left( n-\sqrt{{{n}^{2}}+\frac{n}{2}+1} \right) \right)\] |
| \[=\underset{n\to \infty }{\mathop{\lim }}\,\sin \left[ \frac{\pi \left( {{n}^{2}}-{{n}^{2}}-\frac{n}{2}-1 \right)}{n+\sqrt{{{n}^{2}}+\frac{n}{2}+1}} \right]\] |
| \[=\underset{n\to \infty }{\mathop{\lim }}\,\sin \left[ \frac{\pi \left( -\frac{1}{2}-\frac{1}{n} \right)}{1+\sqrt{1+\frac{1}{2n}+1}} \right]\] |
| \[\sin \left( -\frac{\pi }{4} \right)=-\frac{1}{\sqrt{2}}\] |
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