JEE Main & Advanced Mathematics Differentiation Question Bank Self Evaluation Test - Limits and Derivatives

  • question_answer
    The value of  \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,{{\left[ {{1}^{1/{{\cos }^{2}}x}}+{{2}^{1/{{\cos }^{2}}x}}+...+{{n}^{1/co{{s}^{2}}x}} \right]}^{{{\cos }^{2x}}}}\] is

    A) 0

    B) n

    C) \[\infty \]

    D) \[\frac{n(n+1)}{2}\]

    Correct Answer: B

    Solution :

    [b] \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,{{\left[ {{1}^{1/{{\cos }^{2}}x}}+{{2}^{1/{{\cos }^{2}}x}}+...+{{n}^{1/{{\cos }^{2}}x}} \right]}^{{{\cos }^{2}}x}}\] \[=\underset{t\to \infty }{\mathop{\lim }}\,{{({{1}^{t}}+{{2}^{t}}+...+{{n}^{t}})}^{1/t}}\] \[\left[ On\text{ }putting\,\,\frac{1}{{{\cos }^{2}}x}=t\ge 1 \right]\] \[=\underset{t\to \infty }{\mathop{\lim }}\,{{({{n}^{t}})}^{1/t}}{{\left[ {{\left( \frac{1}{n} \right)}^{t}}+{{\left( \frac{2}{n} \right)}^{t}}+...+{{\left( \frac{n}{n} \right)}^{t}} \right]}^{1/t}}\] \[=n\underset{t\to \infty }{\mathop{\lim }}\,{{\left[ {{\left( \frac{1}{n} \right)}^{t}}+{{\left( \frac{2}{n} \right)}^{t}}+...+{{\left( \frac{n}{n} \right)}^{t}} \right]}^{1/t}}\] \[=n{{(0+0+...+1)}^{0}}=n.\]


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