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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    True statement for \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{2+3x}-\sqrt{2-3x}}\] is                 [BIT Ranchi 1982]

    A) Does not exist

    B) Lies between 0 and \[\frac{1}{2}\]

    C) Lies between \[\frac{1}{2}\] and 1

    D) Greater then 1

    Correct Answer: B

    Solution :

    • \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{(1+x)-(1-x)}{(2+3x)-(2-3x)}\,\left[ \frac{\sqrt{2+3x}+\sqrt{2-3x}}{\sqrt{1+x}+\sqrt{1-x}} \right]\]                    \[=\frac{1}{3}\,\left[ \frac{2\sqrt{2}}{2} \right]=\frac{\sqrt{2}}{3},\,\,0<\frac{\sqrt{2}}{3}<\frac{1}{2}.\]           
    • Aliter : Apply L-Hospital?s rule,             
    • \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{2+3x}-\sqrt{2-3x}}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}}{\frac{3}{2\sqrt{2+3x}}+\frac{3}{2\sqrt{2-3x}}}\]                              
    • \[=\frac{\frac{1}{2}+\frac{1}{2}}{\frac{3}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}}=\frac{2\sqrt{2}}{6}=\frac{\sqrt{2}}{3}\].


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