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Math Help - differentiability at point

  1. #1
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    differentiability at point

    consider the function: f(x) = x^2cos(1/x) if x \neq 0. 0 if x=0. show f is differentiable at 0? is f continuous at 0? if f'(x) continuous at 0?. ... ok that is the problem. but i dont know how to take the limit since it is cos(1/h) and then you cant just stick in the zero. i cant use l'hopital which is what i usually try. using the limit of differentiability i get the derivative is hcos(1/h) as h goes to 0. but i cant do anything with it.
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    Re: differentiability at point

    Quote Originally Posted by learning View Post
    using the limit of differentiability i get the derivative is hcos(1/h) as h goes to 0. but i cant do anything with it.
    The limit of differentiability, eh? So, differentiability is a function of which you can take a limit, right? Do you know the value of differentiability at 5?

    Concerning the limit of h\cos(1/h) as h\to0, what happens to a magnitude that is locked between -1 and 1 as you multiply it by smaller and smaller values?
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    Re: differentiability at point

    come on now you ok i did mistype. sorry. i mean limit definition of differentiability (at a point/general. at a point here). as for your remark on this limit.. i know the limit is zero and i know why. but how can i state it to be so rather just by intuition? i get no mark if i just say it is obvious. ok so since the limit of h as h goes to zero, you are multiplying a 0 by .... well anything would make it zero. BUT, the limit of cos(1/h) as h goes to zero doesnt exist so i cant just use product of limits = limit of products.
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    Re: differentiability at point

    Quote Originally Posted by learning View Post
    come on now you ok i did mistype. sorry. i mean limit definition of differentiability (at a point/general. at a point here). as for your remark on this limit.. i know the limit is zero and i know why. but how can i state it to be so rather just by intuition? i get no mark if i just say it is obvious. ok so since the limit of h as h goes to zero, you are multiplying a 0 by .... well anything would make it zero. BUT, the limit of cos(1/h) as h goes to zero doesnt exist so i cant just use product of limits = limit of products.



    f'(a)={\lim _{h \to 0}}\frac{{f(a + h) - f(a)}}{h} and f(x) = \left\{ {\begin{array}{*{20}{c}}{x^2\cos \left( {\frac{1}{x}} \right),}&{x \ne 0}\\{0,}&{x = 0}\end{array}} \right.

    h\ne 0,~\frac{{f(0+ h) - f(0)}}{h}=h\cos \left( {\frac{1}{h} \right)

    -|h|\le\left|h\cos \left( {\frac{1}{h} \right)\right|\le |h|

    Let h\to 0
    Last edited by Plato; March 25th 2013 at 02:41 PM.
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    Re: differentiability at point

    If you have an intuitive understanding why h\cos(1/h) tends to 0 (as a product of a function that tends to 0 and a bounded function), it is easy to write an epsilon-delta proof that the limit is zero. Let's proof a generalization.

    Theorem. Let f(x) be such that f(x)\to0 as x\to x_0 and let g(x) be bounded in some neighborhood of x_0. The latter means that there exist positive constants r and M such that |x - x_0| < r implies |g(x)|\le M for all x. Then f(x)g(x)\to0 as x\to0.

    Proof. By definition of limit, for every \varepsilon>0 there exist a \delta>0 such that 0<|x-x_0|<\delta implies |f(x)|<\varepsilon for all x. Suppose some \varepsilon>0 is given. Fix the corresponding \delta. What happens to |f(x)g(x)|=|f(x)|\cdot|g(x)| when both 0<|x-x_0|<\delta and |x-x_0|<r, i.e., when 0<|x-x_0|<\min(\delta,r)? Then |f(x)g(x)|<M\varepsilon. We would like to make |f(x)g(x)|<\varepsilon to conform to the definition of limit. For this, we choose the \delta that corresponds not to \varepsilon but to \varepsilon/M. Then 0<|x-x_0|<\delta implies |f(x)|<\varepsilon/M and 0<|x-x_0|<\min(\delta,r) implies |f(x)g(x)|<\varepsilon for all x.

    Altogether, given an \varepsilon>0, there exists a \delta', namely, \delta'=\min(\delta,r) where \delta is obtained from the definition of limit of f(x) for \varepsilon/M. For this \delta' it is true that 0<|x-x_0|<\delta' implies |f(x)g(x)|<\varepsilon for all x.

    You are left with an exercise to understand this proof, to apply it to the current problem and simplify where possible for concrete f(x) = x and g(x) = \cos(1/x).
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