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

  • question_answer
    Let \[f(x)=x-[x],\] where [x] denotes the greatest integer \[\le x\] and \[g(x)=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{{{\{f(x)\}}^{2n}}-1}{{{\{f(x)\}}^{2n}}+1},\] then g(x) is equal to

    A) 0

    B) 1

    C) -1

    D) None of these  

    Correct Answer: C

    Solution :

    [c] As \[0\le x-[x]<1\forall x\in R,0\le f(x)<1\] \[\therefore \,\,\,\,\underset{n\to \infty }{\mathop{\lim }}\,{{\{f(x)\}}^{2n}}=0\] Thus, for \[x\in R,\] \[g(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{\{f(x)\}}^{2n}}-1}{{{\{f(x)\}}^{2n}}+1}=\frac{0-1}{0+1}=-1.\]


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