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Math Help - Derivative/limit

  1. #1
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    Derivative/limit

    I am having trouble with this problem. It asks to: Prove that the function f(x)= x^2*sin(1/x) if x doesn't equal zero and 0 if x=0 is differentiable at x=0 and to compute the derivative at zero.


    This is what i've done so far:

    Noted f(0)=0 so its defined at 0.
    Proved that the limit of f(x) as x->0 is 0 through the squeeze theorem.
    -Through this f(x) is continuous*

    I think i need to prove that the limit of the derivative from the left and right as x->0 = some number correct? or am i wrong? this way it would prove no sharp turns and the derivative at 0? Can i get some hints or help? Thank you!
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  2. #2
    Moo
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    Hello,

    For f to be differentiable at x=0, prove that \lim_{h \to 0} ~ \frac{f(h)-f(0)}{h}=\lim_{h \to 0} ~ \frac{h^2 \sin(\tfrac 1h)-0}{h}=\lim_{h \to 0} ~ h \sin(\tfrac 1h) is finite
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  3. #3
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    how do i compute the limit at 0 though?
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  4. #4
    o_O
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    Like you said, squeeze theorem

    Start with: -1 \leq \sin \left(\frac{1}{h}\right) \leq 1
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  5. #5
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    yes, but if h is multiplied through and its value is 0 that means the derivative at 0 is 0 and it isn't....:/
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  6. #6
    o_O
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    Why do you think it isn't?
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  7. #7
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    uhhhhhh i was looking at the wrong graph. I did most of that and the graph didn't look the same so i assumed i was wrong, but i just made a typo in my graph. Thanks hahah
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    Reply

    To prove that a function is differentiable at some point, prove the existence and continuity of partial derivatives (or derivative) at this point.
    You must thus compute the derivative for x not equal to 0 (say a) and the derivative at x = 0 (say b) and prove that limit when x goes to 0 of a = b. Just the way you did it to prove continuity on the function.
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