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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    The value of\[\underset{x\to \infty }{\mathop{\lim }}\,=\frac{{{x}^{2}}\sin \left( \frac{1}{x} \right)-x}{1-|x|}\]is

    A)  \[-1\]             

    B)  \[1\]

    C)  0              

    D)  \[\infty \]

    Correct Answer: C

    Solution :

     \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{2}}\sin (1/x)-x}{1-|x|}\] Let\[x=1/t,\]then \[\underset{t\to 0}{\mathop{\lim }}\,\frac{\frac{1}{{{t}^{2}}}\sin t-\frac{1}{t}}{1-\frac{1}{t}}\] \[=\underset{t\to 0}{\mathop{\lim }}\,\frac{\frac{\sin t}{t}-1}{t-1}=0\]


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