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Homework Help: A question on limits?

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    x^2*{Cos(Pi/x)} when x!=0 (Not equal to)

    0 when x=0
    Is the function differentiable at x=0 ??

    P.S.- Cos(pi/x) is under mod (Absolute value)

    2. Relevant equations - Basic limit formulas.

    3. The attempt at a solution
    I tried to differentiate it wrt to x but I am getting the function to be non differentiable
    However with first principle differentiation I get it as differentiable. Where am I going wrong? Any help appreciated.
  2. jcsd
  3. Apr 16, 2015 #2


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    I am confused as to what you want to show. To show "f is differentiable at x= 0" you want to show that [itex]\lim_{h\to 0} \frac{f(h)- f(0)}{h}[/itex] exits. That is not necessarily dependent on the derivative of f for x not 0.
  4. Apr 16, 2015 #3


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    you should just try to find the derivative at x=0 by applying the definition of derivative as a limit of a ratio. So you basically looking if the limit [itex]\lim\limits_{x \to 0}xcos(\pi/x)[/itex] exists.

    I think what confuses you is that the derivative which in the interval [itex](-\infty,0)+(0,\infty)[/itex] is equal to[itex]2xcos(\pi/x)+sin(\pi/x)[/itex] cannot be defined for x=0, but thats the point we ve excluded 0 in order to be able to do these calculations for the derivative wrt x. For x=0 you just have to follow the definition of derivative as a limit of the ratio.
    Last edited: Apr 16, 2015
  5. Apr 16, 2015 #4


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    The limit [tex]
    \lim_{x \to 0} f'(x) [/tex] does not exist. However the limit [tex]
    f'(0) = \lim_{x \to 0} \frac{f(x)}{x}[/tex] does exist. Thus [itex]f[/itex] is differentiable at 0 but [itex]f'[/itex] is not continuous at 0.
  6. Apr 17, 2015 #5
    That was the exact point of confusion, I think I got it now though thanks.

    This was the exact thing I needed to know, thanks.
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