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Lim {x -> 1} x / (x - 1) = infty

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that [itex]\lim_{x \to 1} \frac{x}{x - 1} = \infty[/itex]

    2. Relevant equations

    3. The attempt at a solution

    First [itex]0 < |x - 1| < \delta[/itex] implies [itex]\frac{x}{x - 1} > N, \forall N[/itex]

    Because the reals are dense, we can choose an [itex]n > 0[/itex] such that [itex]\frac{x}{x - 1} > \frac{n}{x - 1} > N[/itex]

    then [itex]\frac{n}{x - 1} > N[/itex]
    [itex]\frac{1}{x - 1} > \frac{N}{n}[/itex]
    [itex]x - 1 < \frac{n}{N}[/itex]

    So choose [itex]\delta = \frac{n}{N}[/itex]

    and [itex]0 < |x - 1| < \delta[/itex] holds

    --------------------------------------------

    My first question is of course, is this proof correct. Secondly, if it's not, my other question is, is choosing an n as I've done above, a legal operation? Could it be changed to make it a legal operation?

    Thanks,
    - Rob
     
  2. jcsd
  3. Dec 31, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No. Not so good. For one thing the limit is only infinity if x>1 (x approaching 1 from the right). If x approaches from the left it's negative infinity. Start by taking your condition that x/(x-1)>N. Change that into an inequality condition for x. Can you deduce from that how to pick a delta?
     
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