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Homework Help: Definition of Limit

  1. Aug 23, 2008 #1

    1. The problem statement, all variables and given/known data


    1)Prove that > limit(f(x), x = a) = limit(f(a+h), h = 0)

    2. Prove that limit(f(x), x = a) = L iff limit(f(x)-L, x = a) = 0




    2. Relevant equations



    3. The attempt at a solution
    1)I have tried to use the definition |f(x)-limf(a+h)| and |f(a+h)-limf(x)|. But it doesnt seem to be working because no further simplification can be done. I cant find a way to relate it back to |x-2|<delta and |h|<delta because the i cant eliminate the limit of function. I have also tried to assume both limit equals to f(a) and proves them. But again, it stucks.

    2)First I assume that limit(f(x), x = a) = L. Then, the definition follows.
    There exists 0<|x-a|<delta such that |f(x)-L|< epsilon.....(1)
    So, in the case of limit(f(x)-L, x = a) = 0.
    0<|x-a|<delta |f(x)-L-0|=|f(x)-L|<epsilon.... from (1)
    and vice versa.
    But it doesnt look like a proof to me. It more like I am rewriting it in another way. Is there a better way to put it?

    Thanks...
     
  2. jcsd
  3. Aug 23, 2008 #2

    HallsofIvy

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    Science Advisor

    "|f(x)- lim f(a+h)|" is very awkward. The DEFINITION of limit is that if [itex]\lim_{x\rightarrow a}= L[/itex], then, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex] then [itex]|f(x)-L|< \epsilon[/itex]. For [itex]\lim_{h\rightarrow 0}f(x+ h)= L[/itex], using that same definition, you need to show that given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]0< |h- 0|= |h|<\delta[/itex], then [itex]|f(a+ h)- L|< \epsilon[/itex]. What happens if you replace "h" in the second formulation by x- a?

    Rewriting it basically is the proof. The only difference between the first part and the second is that the function f(x) is replace by the function f(x)- L. Okay replace f(x) by that in the definition of limit:
    Given [itex]\epsilon> 0[/itex] there exist [itex]\delta[/itex] such that if [itex]0< |x- a|< \delta[/itex] then [itex]|(f(x)-L)- 0|= |f(x)- L|< \epsilon[/itex]. The fact that |(f(x)-L)- 0| is exactly the same as |f(x)- L| is the whole point.
     
  4. Aug 23, 2008 #3
    For the 1st question, first we assume that
    lim(f(x) , x = a) = L. Then, given any epsilon>0 there exists an delta>0 such that if
    0<|x-a|<delta then |f(x)-L|<epsilon.

    Then, the second part, we assume again lim(f(a+h) , h = 0) = L. Then, given any epsilon>0 there exists an delta>0 such that if
    0<|h|<delta by letting |h|=|x-a| |f(a+h)-L|=|f(x)-L|<epsilon.

    Is this the correct reasoning for the solution? It doesnt seem to be logical to me especially the assumption of lim(f(a+h) , h = 0) = L because if we do that we already assume that the both are equal so the prove is meaningless. But, if I assume lim(f(a+h) , h = 0) = M, it stucks.
     
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