1. Limited time only! Sign up for a free 30min personal 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!

Proving a limit is false when L does not equal 1

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that if [itex]L \neq 1 [/itex], the statement [tex] \lim \limits_{x \to \infty} (1+\frac{1}{x}) = L [/tex] is false.


    2. Relevant equations

    The Definition of a Limit

    3. The attempt at a solution
    So I've been trying to prove this by negating the logical statement of the definition of a limit; i.e. by trying to prove that
    [itex]\exists \epsilon > 0 [/itex] such that [itex]\forall \delta >0 \exists x > \delta [/itex] such that [itex]\left|f(x)-L\right|\geq \epsilon[/itex].
    I know that when [itex]L=1[/itex] the limit exists; that is no trouble to prove. The problem is that every time i try to find an x that works, I can never make it work in my proof. Am I going about this the right way?
     
  2. jcsd
  3. Sep 29, 2011 #2
    Use the limit Law to break up your limit
     
  4. Sep 29, 2011 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Think about it this way. Suppose L > 1. Now you know how to get the function very close to 1, so you should be able to keep it away from L, no?. So think about what would happen if you choose epsilon half the distance from L to 1. And the case L < 1 is similar or you could combine them.
     
  5. Sep 29, 2011 #4
    So if I choose [itex]\epsilon = \frac{L-1}{2}[/itex], I want to find an x that will give me [tex]\left|f(x)-L\right|=\left|1+\frac{1}{x}-L\right|=\epsilon=\frac{L-1}{2}.[/tex]

    This is kind of where I'm stuck. I'm not sure what the best way to manipulate the absolute value sign. I know I can do it with either the triangle or reverse triangle inequality, but which would be the right direction to take? would it matter?
     
  6. Sep 29, 2011 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Use |L-1|/2, in case L < 1.
     
  7. Sep 30, 2011 #6
    ok, so starting with [tex]\left|1+\frac{1}{x}-L\right|\geq\frac{\left| L-1 \right|}{2},[/tex]
    I get
    [tex]\left| 1 \right| + \left| \frac{1}{x} - L \right| \geq \frac{\left| L-1 \right|}{2} \text{(triangle inequality)}[/tex]

    [tex]\left| \frac{1}{x} - L \right| \geq \frac{\left| L-1 \right|}{2} -1 = \frac{\left| L-1 \right|-\left|2\right|}{2} \geq \frac{\left| L-3 \right|}{2} \text{(reverse triangle ineq.)}[/tex]

    So now I'm stuck here at
    [tex] \left| \frac{1}{x} - L \right| \geq \frac{\left| L-3 \right|}{2} .[/tex]

    I'm not too experienced with doing algebra with absolute value signs. Have I been doing it right so far?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving a limit is false when L does not equal 1
Loading...