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Can the limit be taken for any fn?

  1. May 29, 2010 #1
    Hi all,

    I have heard that it is always possible to find the derivative of a function, whatever it is. Since taking the limit of a fn seems to be based on derivatives, does this mean that it is always possible to find a closed-form expression for the limit of a function f(x,..,..,...) as x->x_limit?

  2. jcsd
  3. May 29, 2010 #2
    Hmm, I can't answer your question, but I can say that the derivative of a function doesn't always exist.

    f(x) = (arccoth(x)) (arctanh(x))

    You may think you could take the derivative of each function by the product rule, but the domain of the function doesn't exist, so whatever you get as the result of the derivative would be false.
  4. May 30, 2010 #3


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    No, in a very specific sense "almost all" functions do not have derivatives. you may be thinking, rather, that if a function is differentiable, then it is always possible to use the definition in terms of the "difference quotient limit" to find the derivative. That might be extremely difficult, but theoretically, its true.

    As for "taking the limit of a fn seems to be based on derivatives", I have no idea where you got that impression. It is, in fact, the other way around. Finding derivatives is bases on taking the limit.
  5. May 30, 2010 #4
    Ah OK. I was basing the assumption that limits were based on derivatives because of l'Hoptial's rule.

    Let me ask another question on limits... I know the following is true...

    \int_{x=0}^{1} f(x) \, \mathrm{d}x = \Big[\int f(x) \, \mathrm{d}x\Big]_{x=1} - \Big[\int f(x) \, \mathrm{d}x\Big]_{x=0}

    So consider that say the first of these parts on the RHS yields 0/0, could we take the limit? i.e. are we allowed to write...

    \int_{x=0}^{1} f(x) \, \mathrm{d}x = \mathrm{lim}_{x \rightarrow 1} \Big[\int f(x) \, \mathrm{d}x\Big] - \mathrm{lim}_{x \rightarrow 0} \Big[\int f(x) \, \mathrm{d}x\Big]
  6. May 31, 2010 #5
    Can I also ask, how does one take the limit to the RHS or LHS of a jump discontinuity?

  7. Jun 1, 2010 #6


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    By ignoring the formula for the function on the other side! Exactly how one takes a limit depends strongly on the form of the function.

    For example
    [tex]f(x)= \{\begin{array}{c} 2x+ 1 for x< 0 \\ -3x+ 2 for x> 0\end{array}[/tex]

    has limit from below [itex]\lim_{x\to 0^-} f(x)= \lim_{x\to 0}2x+ 1= 1[/itex], and limit from above, [itex]\lim_{x\to 0^+} f(x)= \lim_{x\to 0} -3x+ 2= 2[/itex]
    Last edited by a moderator: Jun 1, 2010
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