Formal Definition of a Limit at a Point

Key Questions

  • Answer:

    See below

    Explanation:

    The definition of limit of a sequence is:

    Given #{a_n}# a sequence of real numbers, we say that #{a_n}# has limit #l# if and only if

    #AA epsilon>0, exists n_0 in NN // AAn>n_0 rArr abs(a_n-l))< epsilon#

  • Before writing a proof, I would do some scratch work in order to find the expression for #delta# in terms of #epsilon#.

    According to the epsilon delta definition, we want to say:

    For all #epsilon > 0#, there exists #delta > 0# such that
    #0<|x-1|< delta Rightarrow |(x+2)-3| < epsilon#.

    Start with the conclusion.

    #|(x+2)-3| < epsilon Leftrightarrow |x-1| < epsilon#

    So, it seems that we can set #delta =epsilon#.

    (Note: The above observation is just for finding the expression for #delta#, so you do not have to include it as a part of the proof.)

    Here is the actual proof:

    Proof

    For all #epsilon > 0#, there exists #delta=epsilon > 0# such that
    #0<|x-1| < delta Rightarrow |x-1|< epsilon Rightarrow |(x+2)-3| < epsilon#

  • Precise Definitions

    Finite Limit
    #lim_{x to a}f(x)=L# if
    for all #epsilon>0#, there exists #delta>0# such that
    #0<|x-a|< delta Rightarrow |f(x)-L| < epsilon#

    Infinite Limits
    #lim_{x to a}f(x)=+infty# if
    for all #M>0#, there exists #delta>0# such that
    #0<|x-a|< delta Rightarrow f(x)>M#

    #lim_{x to a}f(x)=-infty# if
    for all #N<0#, there exists #delta>0# such that
    #0<|x-a|< delta Rightarrow f(x) < N#

Questions