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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    If \[{{x}_{1}}=3\]and\[x>0\]then \[\underset{n\to \infty }{\mathop{\lim }}\,{{x}_{n}}\] is equal to

    A) -1

    B) 2

    C) \[\sqrt{5}\]

    D) 3

    Correct Answer: B

    Solution :

    • We have \[{{x}_{1}}=3,\,\,{{x}_{n+1}}=\sqrt{2+{{x}_{n}}}\]           
    • \[{{x}_{2}}=\sqrt{2+{{x}_{1}}}=\sqrt{2+3}=\sqrt{5}\], \[\,{{x}_{3}}=\sqrt{2+{{x}_{2}}}=\sqrt{2+\sqrt{5}}\]  \[\therefore \,\,\,{{x}_{1}}>{{x}_{2}}>{{x}_{3}}\]          
    • It can be easily shown by mathematical induction that the sequence \[{{x}_{1}},\,\,{{x}_{2}},........{{x}_{n}},....\] is a monotonically decreasing sequence bounded below by 2. So it is convergent. Let \[\lim {{x}_{n}}=x.\]
    • Then  \[{{x}_{n+1}}=\sqrt{2+{{x}_{n}}}\,\Rightarrow \,\,\lim {{x}_{n+1}}=\sqrt{2+\lim {{x}_{n}}}\]\[\Rightarrow \,x=\sqrt{2+x}\]           
    • \[\Rightarrow \,\,{{x}^{2}}-x-2=0\,\,\Rightarrow \,\,(x-2)\,(x+1)=0\,\Rightarrow \,x=2\]                                                                      
    • \[(\because \,\,{{x}_{n}}>0\,\forall \,n;\,\,\therefore \,\,x>0)\]


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