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JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    The value of \[\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{2}}+3x+2}{{{x}^{2}}+4x+3}\]is equal to [Pb. CET 2000]

    A)                 0

    B)                 1

    C)                 2

    D)                 ½

    Correct Answer: D

    Solution :

                       \[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}+3x+2}{{{x}^{2}}+4x+3}=\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{2}}+2x+x+2}{{{x}^{2}}+3x+x+3}\]                                 \[=\underset{x\to -1}{\mathop{\lim }}\,\frac{(x+1)(x+2)}{(x+1)(x+3)}=\underset{x\to -1}{\mathop{\lim }}\,\frac{x+2}{x+3}=\frac{1}{2}\].


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