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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    \[\underset{n\to \infty }{\mathop{\lim }}\,\sin [\pi \sqrt{{{n}^{2}}+1}]=\]        

    A) \[\infty \]

    B) 0

    C) Does not exist

    D) None of these

    Correct Answer: B

    Solution :

    • Given limit \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\sin \left\{ n\pi {{\left( 1+\frac{1}{{{n}^{2}}} \right)}^{1/2}} \right\}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\sin \,\left\{ n\pi \left( 1+\frac{1}{2{{n}^{2}}}-\frac{1}{8{{n}^{4}}}+... \right) \right\}\]                         
    • \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\frac{1}{3}\,{{(27-2x)}^{-2/3}}(-2)}{-\frac{3}{5}\,{{(243+5x)}^{-4/5}}(5)}=2.\]                                                                    
    • \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,{{(-1)}^{n}}\,\sin \pi \,\left( \frac{1}{2n}-\frac{1}{8{{n}^{3}}}+.... \right)=0.\]


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