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Definition of Limit

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that lim (f(x) , x=a) = a4 where f(x)=x4


    2. Relevant equations
    I think I understand the mechanics of limit proof. I just want to improve my way of presenting it because sometimes even I feel ambiguous for my own works.


    3. The attempt at a solution
    For all [tex]\epsilon[/tex]>0, there exists [tex]\delta[/tex]>0 such that if 0 < |x-a| < [tex]\delta[/tex], then |f(x)-a4| < [tex]\epsilon[/tex]

    I think the mechanic of the proof is I need to find delta in terms of epsilon such that it satisfies all the conditions above. (or another way round, for me, the proof more like I need to express f(x)-L where L is the limit in terms of delta so that I could make it smaller than epsilon... please correct me)

    So, the first thing I need to do is to express |f(x)-a4| in terms of delta.
    |f(x)-a4|
    =|x^4 - a^4|
    =|x-a| * |x3 + ax2 + a2x + a3|
    <=|x-a| * |x3| + |ax2| + |a2x| + |a3|

    My problem comes, (when it starts to sound ambiguous)
    What I do is I will let |x-a| < [tex]\delta[/tex]1 where [tex]\delta[/tex]1 is a positive number such that |x3| + |ax2| + |a2x| + |a3| < P where P is a positive number.

    My reason of doing this is I need to obtain a close interval for the expression |x3| + |ax2| + |a2x| + |a3| so that I can make |f(x) - L| smaller than epsilon. Therefore, I started out by letting |x-a| smaller than [tex]\delta[/tex]1 which is a arbitrary positive number. Then by algebraic manipulation, I can obtain an inequality of |x| smaller than [tex]\delta[/tex]1 +a . So when I substitute [tex]\delta[/tex]1 +a into each terms of |x3| + |ax2| + |a2x| + |a3|, I can get inequality of each terms smaller than something. To save the trouble, I let the whole expression less than P which is also an arbitrary positive number.

    Since |x3| + |ax2| + |a2x| + |a3| < P
    then |x-a| * |x3| + |ax2| + |a2x| + |a3| < [tex]\delta[/tex] *P

    So, when I let [tex]\delta[/tex] *P = [tex]\epsilon[/tex]. (I should express delta in terms of epsilon or epsilon in terms of delta?? The first sounds more correct for me because our mission is to find a delta for any given epsilon)

    |f(x)-a4| < [tex]\epsilon[/tex]

    Done.

    Please feel free to correct me.
     
  2. jcsd
  3. Aug 26, 2008 #2
    I'll rewrite what you've wrote thus far.

    [tex]\forall \epsilon > 0, \exists \delta > 0[/tex] such that [tex] \left| x^4 - a^4 \right| < \epsilon[/tex] whenever [tex]0 < \left| x -a \right| < \delta[/tex]

    From here, we need:

    [tex]\left| x^4 - a^4 \right| < \epsilon [/tex]

    Factor we get:

    [tex] \left| x - a \right| \left| x^3 + a^{2}x+x^{2}a+a^3 \right| < \epsilon[/tex]

    Since we have that [tex] \delta > \left| x - 2 \right| [/tex], then you need to find a delta such that

    [tex] \left| x - a \right| \left| x^3 + a^{2}x+x^{2}a+a^3 \right| < \left| x - a\right|*P < \delta*P < \epsilon[/tex]

    Naturally, you want such that [tex]\delta = \frac{\epsilon}{P}[/tex]

    Hope that helps.
     
  4. Aug 27, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You need parentheses, of course. But there is no reason to break up |x3+ ax2+ a2+ a3|. It's awkward and doesn't help.

    You are assuming, in any case, that x is close to a, say a-1< x< a+1. x2< (a+1)2 and x3< (a+ 1)3. That should let you set an upperbound on |x3+ ax2+ a2+ a3|.
     
  5. Aug 27, 2008 #4
    It helps a lot. Thank you.

    I have a paradox here. Please tell me what is wrong.
    I need to prove that limit (f(x) , x = a-) = -[tex]\infty[/tex]
    f(x) = [tex]\frac{x}{(x-1)^2(x-3)}[/tex]

    1st case

    For all M , where M is a arbitrary large number, there exists [tex]\delta[/tex]>0 such that f(x)<M whenever 0 < a-x < [tex]\delta[/tex]

    0 < a-x < [tex]\delta[/tex].
    a-[tex]\delta[/tex] < x < a
    Therefore x<a ....(1)

    Let |x-a| < [tex]\delta[/tex]1 where [tex]\delta[/tex]1 is a positive number such that [tex]\frac{1}{(x-1)^2(x-3)}[/tex] < P where P is a positive number. ...(2)

    f(x) = [tex]\frac{x}{(x-1)^2(x-3)}[/tex]
    < [tex]\frac{a}{P}[/tex] .... from (1) and (2)
    Let [tex]\frac{a}{P}[/tex]=M
    f(x) < M

    Done.

    2nd case

    For all M , where M is a arbitrary large number, there exists [tex]\delta[/tex]>0 such that f(x)>M whenever 0 < a-x < [tex]\delta[/tex]
    I proved this for fun at first. But, it bothers me when it became confusing that the proof contradicts itself.

    As above,
    0 < a-x < [tex]\delta[/tex].
    a-[tex]\delta[/tex] < x < a
    Therefore x>a-[tex]\delta[/tex] ....(1)

    Let |x-a| < [tex]\delta[/tex]1 where [tex]\delta[/tex]1 is a positive number such that [tex]\frac{1}{(x-1)^2(x-3)}[/tex] > P where P is a positive number. ...(2)

    f(x) = [tex]\frac{x}{(x-1)^2(x-3)}[/tex]
    > [tex]\frac{a-\delta}{P}[/tex] .... from (1) and (2)
    Let [tex]\frac{a-\delta}{P}[/tex]=M
    f(x) > M

    Done.

    I derive the both cases with the same arguments but obtain opposite result. What is wrong with my proof and how valid this proof can be?

    How to type limit notation by LATEX...
     
    Last edited: Aug 27, 2008
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