JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}\]is equal to:

    A) \[\lambda \]                                      

    B) \[-1\]

    C)  zero                                     

    D)  does not exist

    Correct Answer: D

    Solution :

    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}\]                 \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-(1-2{{\sin }^{2}}x)}}{\sqrt{2}x}\]                 \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{2{{\sin }^{2}}x}}{\sqrt{2}x}\]                 \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{|\sin x|}{x}\] Let                \[f(x)=\frac{|\sin x|}{x}\] Now,     \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|\sin (0+h)|}{0+h}\]                                  \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin h}{h}=1\] and        \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|\sin (0-h)|}{-h}=-1\] The limit of function does not exist.


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