JEE Main & Advanced Sample Paper JEE Main - Mock Test - 12

  • question_answer
    The value of \[\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{x+\sqrt{x+\sqrt{x+...+\sqrt{x}}}}\] is

    A) \[0\]

    B) \[2\]

    C) \[1\]

    D) \[1/2\]

    Correct Answer: D

    Solution :

    Let, \[y=\underset{x\to \infty }{\mathop{\lim }}\,\,\,\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+}}}}\]
    \[\Rightarrow \,\,\,y=\underset{x\to \infty }{\mathop{\lim }}\,\,\,\sqrt{x+y}\]
    \[\Rightarrow \,\,\,{{y}^{2}}=x+y\]
    \[\Rightarrow \,\,\,{{\left( y-\frac{1}{2} \right)}^{2}}=\left( x+\frac{1}{4} \right)\]
    \[\therefore \] vertex of parabola is \[\left( -\frac{1}{4},\frac{1}{2} \right)\]
    Now, range of function i.e., \[y=\underset{x\to \infty }{\mathop{\lim }}\,\,\,f(x)=\frac{1}{2}\]


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