1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of Limit Paradox

  1. Aug 28, 2008 #1
    I have a paradox here. Please tell me what is wrong.
    I need to prove that [tex]\lim_{x \rightarrow a-}f(x) = -\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 (The opposite 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]

    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?
     
  2. jcsd
  3. Aug 28, 2008 #2
    Haven't read yoru question but you would do \lim_{x \rightarrow a} to give you:

    [tex] \lim_{x \rightarrow a}[/tex]
     
  4. Aug 28, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    There's a LOT wrong that 'proof'. It doesn't really prove anything. You can't even BEGIN to prove something about the limit of f(x) as x->a until you know what a is!!!! What is it?
     
  5. Aug 28, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    One problem you have is that you haven't said what a is!
    [tex]\lim_{x \rightarrow a-}f(x) = -\infty[/tex]
    is true only if a= 1. where did you use that?
     
  6. Aug 28, 2008 #5
    You all are right. It is impossible to define the limit if a is not given.
    I confuse this with the case [itex]\lim_{x \rightarrow a} f(x) = a^4[/itex],
    where f(x) = x4.

    https://www.physicsforums.com/showthread.php?t=252170

    I think it is correct in x4 case because it is continuous and the limit of it when x is approaching "a" is always equals to a4 whereas in this question, I can only define that when x is approaching a certain "a", the limit is -[tex]\infty[/tex] but generally when a is any numbers, the limit is not [tex]-\infty[/tex].

    Please help me to check this so that I can make sure that what I am doing is correct.
    Prove that [itex]\lim_{x \rightarrow 1}f(x) = 1[/itex] , where [itex]f(x) = \frac{1}{x}[/itex].

    For all \epsilon >0 , there exists \delta >0 , such that whenever 0<|x-1|<\delta , then [itex]|f(x)-1|<\epsilon[/itex].

    |f(x)-1|
    [itex]= |\frac{1}{x} - 1|[/itex]
    [itex]= |\frac{1-x}{x}|[/itex]
    [itex]=\frac{|1-x|}{|x|}[/itex]
    [itex]=|1-x|*\frac{1}{x}[/itex]

    By letting [itex]\delta = \delta_1[/itex] where [itex]\delta_1[/itex] is an arbitrary positive number such that [itex]\frac{1}{x} < M[/itex] , where M is an arbitrary positive number. (This is the part I am not very sure, my reasoning is there exists [itex]0<|x-1|<\delta_1/[/itex] such that I can get [itex]\frac{1}{|x|} < M[/itex], but how can I make sure that this assumption is always correct)

    So, [itex]|f(x)-1| < \delta*M = \epsilon[/itex]
    where [itex]\delta = min [ \delta_1 ,\frac{M}{\epsilon}][/itex]
     
    Last edited by a moderator: Aug 29, 2008
  7. Aug 28, 2008 #6
    Y some of the LATEX is not displayed?
     
  8. Aug 29, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Because you didn't tell it to! You did not include the [ tex ] and [ / tex ] tags. Also do not use the HTML tags and in LaTex. use ^ and _ instead.
     
  9. Aug 29, 2008 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You need to be a little more careful. If, say, 0< |x-1|< 1/2, then -1/2< x- 1< 1/2 so
    1/2< x< 3/2. Now you can say that 1/x< 1/(1/2)= 2. |f(x)-1|< 2|1- x|.
     
  10. Aug 29, 2008 #9
    Consider the function f(x) = x/{(x-1)2(x-3)}

    If x has any value other than 1 or 3, the value of f(x) can be calculated by substituting the value of x in the formula. For example, if x=5, f(x) = f(5) = 5/(42*2) = 5/32. The limit of f(x), for x approaching 5, whether from above or from below, will also be 5/32, as f is a continuous function on any interval that does not include 1 or 3.

    The only nontrivial limits are therefore those for x approaching 1 or 3 from above or from below, and those for x approaching plus or minus infinity (that is, x increasing or decreasing without bound).

    First we should realize that f(x) = 0 if and only if x = 0.

    For very large |x|, f(x) will be nearly equal to x/x3 = 1/x2, which approaches zero from above if x approaches plus or minus infinity. The same should be true for f(x).

    For x approaching 1 from below (i.e. x<1 and x<3, but x>0) f(x) will have negative values.
    For x approaching 1 from above (i.e. x>1 but x<3) f(x) will also have negative values.
    For x approaching 3 from below (i.e. x<3, but x>1) f(x) will still have negative values.
    For x approaching 3 from above (i.e. x>3 and x>1) f(x) will have positive values.

    If |x-1| < epsilon, |f(x)| will be larger than 1/(3-1)*1/(epsilon)2.
    If |x-3| < epsilon, |f(x)| will be larger than 1/(3-1)2*1/epsilon.
    These absolute values will therefore increase without bounds when epsilon approaches zero.

    Therefore f(x) approaches minus infinity if x approaches 1 from above or below, or 3 from below; and f(x) approaches plus infinity if x approaches 3 from above.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Definition of Limit Paradox
  1. Definition of Limit (Replies: 2)

  2. Definition of Limit (Replies: 3)

  3. Definition of Limit (Replies: 2)

  4. Definition of a limit (Replies: 3)

  5. Definition of a Limit (Replies: 1)

Loading...