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What is the difference between differentiability and continuity at a point?

  1. Oct 9, 2012 #1
    Could someone explain this to me in terms of limits and derivatives instead of plain english?

    For example, how would you solve a question that says

    find whether the function f is differentiable at x=n

    and a question that asks

    find whether the function f is continuous at = n



    thanks for any help
     
  2. jcsd
  3. Oct 9, 2012 #2

    ehild

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  4. Oct 9, 2012 #3

    HallsofIvy

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    The function y= |x| is continuous at x= 0 but not differentiable there. Look at its graph.
    It is a basic theorem that if a function is differentiable at a point, then it is continuous that but the other way around is not true.

    Essentially, saying that a function is continuous means that its graph has no "breaks". Saying that a function is differentiable means that its graph has no "corners" as well.

    Oops! That was "plain English" and you said "in terms of limits and derivatives instead of plain english".

    A function, f, is "continuous at x= n" if and only if
    1) f(n) exists
    2) [itex]\lim_{x\to n} f(x)[/itex] exits
    3) [itex]\lim_{x\to n} f(x)= f(n)[/itex]
    (since (3) wouldn't make sense if the two sides didn't exist, we often just cite it)

    A function, f, is "differentable at x= n" if and only if
    [tex]\lim_{h\to 0} \frac{f(n+h)- f(n)}{h}[/tex]
    exists.
     
    Last edited: Oct 9, 2012
  5. Oct 9, 2012 #4
    But this is not enough to satisfy continuity. To be differentiable, the function has to also be continuous (or equivalently (?), differentiating from the left must equal differentiating from the right).

    [tex]\lim_{h\to 0} \frac{f(n+h)- f(n)}{h} = \lim_{h\to 0} \frac{f(n)- f(n-h)}{h} [/tex]
     
  6. Oct 9, 2012 #5

    HallsofIvy

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    ??? Why are you assuming that I was requiring h to be positive? "[itex]\lim_{h\to 0}[/itex]", as opposed to "[itex]\lim_{h\to 0^+}[/itex]" or "[itex]\lim_{h\to 0^-}[/itex]" is the "two sided limit".
     
    Last edited: Oct 9, 2012
  7. Oct 9, 2012 #6

    Ray Vickson

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    This is already implied in the original definition, because that definition did not restrict the sign of h.

    RGV
     
  8. Oct 9, 2012 #7

    jbunniii

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    Differentiability IS enough to satisfy continuity. To see this, suppose that [itex]f[/itex] is differentiable at [itex]n[/itex]. Note that if [itex]h \neq 0[/itex], then
    [tex]f(n+h) - f(n) = \frac{f(n+h) - f(n)}{h} \cdot h[/tex]
    Taking limits of both sides:
    [tex]\lim_{h \rightarrow 0} \left(f(n+h) - f(n)\right) = \lim_{h \rightarrow 0} \left(\frac{f(n+h) - f(n)}{h} \cdot h\right) = \left(\lim_{h \rightarrow 0} \frac{f(n+h) - f(n)}{h}\right) \cdot \left(\lim_{h \rightarrow 0} h\right) = f'(n) \cdot 0 = 0[/tex]
    Therefore [itex]\lim_{h \rightarrow 0} f(n+h) = f(n)[/itex], so [itex]f[/itex] is continuous at [itex]n[/itex].
     
  9. Oct 10, 2012 #8
    :D oops! Disregard that, I don't know anything about mathematics!
     
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