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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{2x-1}{\sqrt{{{x}^{2}}+2x+1}}\]is equal to

    A)  2                            

    B)         \[-2\]

    C)  1                            

    D)         \[-1\]

    E)  0

    Correct Answer: B

    Solution :

    \[\underset{x\to -\infty }{\mathop{\lim }}\,=\frac{2x-1}{\sqrt{{{x}^{2}}+2x+1}}=\underset{y\to \infty }{\mathop{\lim }}\,\frac{-2-\frac{1}{y}}{\sqrt{1-\frac{2}{y}+\frac{1}{{{y}^{2}}}}}\] \[(Putx=-y\therefore x\to -\infty \,ie,y\to \infty )\] \[=-\frac{1}{2}=-2\]


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