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Limit Function Problem

  1. Aug 20, 2011 #1

    I am currently trying to understand some limits. |x| is not differentiable because left and right limits are different.

    I also learned that |x|/x=sgn(x) (signum)


    [tex]\lim_{\substack{x\rightarrow 0}} sgn(x)[/tex]

    is not defined, ergo function abs(x) is not differentiable in 0.

    But isn't sgn(x) defined to be 0 if x=0, or am I misinterpreting the term "approaching".

    Another example:


    [tex]g(y) = \begin{Bmatrix}
    1, & y \not= 0 \\ 0, & y=0 \end{Bmatrix} [/tex]

    and f(x)=0 for all x.

    [itex]\lim_{\substack{y\rightarrow 0}} g(y)=1[/itex]

    [itex]\lim_{\substack{x\rightarrow 0}} f(x)=0[/itex]


    [itex]\lim_{\substack{x\rightarrow 0}} (g \circ f) (x) =0 [/itex]

    This is example where composition of functions doesn't work as they should.

    I don't understand how this composition of functions, doesn't work. Can anybody explain to me how this works? ( I found the example on the internet, but not as good explanation)
    Last edited: Aug 20, 2011
  2. jcsd
  3. Aug 20, 2011 #2


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    For the first question: lim sgn(x), = 1 for x-> 0 and x > 0, = -1 for x ->0 and x < 0. Therefore the limit does not exist. Defining sgn(0)=0 or anything else doesn't matter.

    For the second [I'm assuming you are asking about g(f(x))]: f(x)=0 for all x, therefore this function is g(0)=0 for all x. I am not quite sure what "doesn't work".
  4. Aug 20, 2011 #3
    It doesn't work like this:

    [tex]\lim_{\substack{x\rightarrow a}} f(x)=b [/tex]
    [tex]\lim_{\substack{y\rightarrow b}} g(y)=c [/tex]

    [tex]\lim_{\substack{x\rightarrow a}} (g \circ f) (x) =c [/tex]

    This feels right to me, But i cannot understand the problem above.(2nd one)

    Does that "for all x" makes everything in composed function 0?
  5. Aug 21, 2011 #4


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    The problem (why it doesn't work) is simply the discontinuity at x=0 for g. For the composite function the argument for g is always 0. The general property for the limit involving g assumes -> 0 through non-zero values of the argument.
  6. Aug 22, 2011 #5


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    sgn(x)= 1 if x> 0, sgn(0)= 0, and sgn(x)= -1 if x< 0.

    In particular,
    [tex]\lim_{x\to 0^+} sgn(x)= \lim_{x\to 0} 1= 1[/tex]
    [tex]\lim_{x\to 0^-} sgn(x)= \lim_{x\to 0} -1= -1[/tex]

    Of course, [itex]\lim_{x\to a} f(x)= L[/itex] if and only if both [itex]\lim_{x\to a^+} f(x)= L[/itex] and [itex]\lim_{x\to a^-} f(x)= L[/itex]. If the two sided limits are not the same, then the limit itself cannot exist.

    As for your second problem, in general [itex]\lim_{x\to a} g(f(x))= g(\lim_{x\to a}f(x))[/tex] if and only if g is continuous at [itex]\lim_{x\to a}f(x)[/itex]. In fact, that can be used as a definition of "continuous". Here, you function g is not continuous at x= 0.

    That's why the limit at 0 does not exist. It has nothing to do with the value of the function at x= 0.
  7. Aug 22, 2011 #6
    My brain is swelling of information i got recently. I misplaced the definition of "not continuous". Thank you.
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