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Differentiation Watch

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    Consider the function f(x) = \left\{\begin{array}{lr}x^{3}\co  s(\frac{1}{x^{2}}) & : x \neq 0\\

       0 & : x = 0

     \end{array}

   \right.

    (i) Calculate f'(x) for  x \neq 0 and show that lim_{x \to 0}f'(x) does not exist. Done
    (ii) Show, by definition of the derivative that f is differentiable at x = 0 and the derivative of f there. Done I think
    (iii) Use this to answer the question: If a function has derivative defined at all points, is that derivative necessarily continuous? Done but depends on part 2
    For the second part it says from the definition of the derivative so we use f'(x) = lim_{h \to 0}\frac{f(x + h) - f(x)}{h} to get:

    lim_{h \to 0} \frac{(x + h)^{3}\cos(\frac{1}{(x + h)^{2}}) - x^{3}\cos(\frac{1}{x^{2}})}{h}

    if x = 0:

    lim_{h \to 0} \frac{h^{3}\cos(\frac{1}{h^{2}})  }{h} = lim_{h \to 0} h^{2}\cos(\frac{1}{h^{2}})

    Can I now say that if h goes to 0, the undefined fraction doesn't matter as the answer is always 0 (due to being multiplied by 0)?

    Part 3:
    If my answer for part 2 is correct then no the derivative doesn't have to be continuous?
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    For that last line you have, cos(a) is always in the interval [-1,1], so it's perfectly safe to say that h^2 cos(\frac{1}{h^2}) -> 0
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    (Original post by marcusmerehay)
    For that last line you have, cos(a) is always in the interval [-1,1], so it's perfectly safe to say that h^2 cos(\frac{1}{h^2}) -> 0
    Woop. Cheers!
 
 
 
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