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Math Help - Derivative proof

  1. #1
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    Derivative proof

    Problem. Let g(x)=x^{2}sin( \frac {1}{x} ), find g'(0) and prove that g'(x) is not continuous at x=0.

    My work:

    g'(0) = \lim _{x \rightarrow 0 } \frac {g(x)-g(0)}{x-0} = \lim _{x \rightarrow 0 } \frac {x^{2}sin( \frac {1}{x} )-g(0)}{x}, but g(0) is undefined, so how do you find the derivative?

    Thanks.
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  2. #2
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    No, it is defined g(0)=0.
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  3. #3
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    Well, then, I have  \lim _{x \rightarrow 0 } \frac {x^{2}sin ( \frac {1}{x} )}{x} = \lim _{x \rightarrow 0 } xsin( \frac {1}{x} )

    So g'(0) = 0?
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  4. #4
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    Yes because \left| x\sin \frac{1}{x} \right| \leq |x| now use squeeze theorem.
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  5. #5
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    Now to prove g'(x) is not continuous at x = 0.

    Pick a sequence  \{ x_{n} \} that converges to 0.

    Consider  \lim _{n \rightarrow \infty } | g'( x_{n}) - 0 | = \lim _{n \rightarrow \infty } | \lim _{x \rightarrow x_{n}} \frac {g(x) - g(x_{n})}{x-x_{n}} |

    I think I'm stuck...
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  6. #6
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    To show that \sin \frac{1}{x} is not continous at zero (for example) a sequene say x_n = \frac{1}{\pi n} which is zero. And pick another sequence like x_n = \frac{1}{\frac{\pi}{2}+\pi n} which is 1. So it is not continous.
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