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Limit Proof

  1. Jul 6, 2010 #1


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    So, I'm trying to prove that [itex]\lim_{x \to a} f(x) = \lim_{h \to 0} f(a+h)[/itex] and I'm wondering if this "proof" is correct:

    Suppose that [itex]\lim_{x \to a} f(x) = l[/itex] and [itex]\lim_{h \to 0} f(a+h) = m[/itex] with [itex]l \neq m[/itex]. Then, for any [itex]\varepsilon > 0[/itex] there must be some [itex]\delta > 0[/itex] such that [itex]0 < |x-a| < \delta[/itex] implies that [itex]|f(x) - l| < \varepsilon[/itex] and [itex]0< |h| < \delta[/itex] that [itex]|f(a+h) - m| < \varepsilon[/itex].

    Clearly [itex]a - \delta < x < a + \delta[/itex] and [itex]a - \delta < a + h < a + \delta[/itex], so there is a number [itex]x_1[/itex] such that [itex]|f(x_1) - l| < \varepsilon[/itex] and [itex]|f(x_1) - m| < \varepsilon[/itex].

    Choosing [itex]\varepsilon = |l-m|/2[/itex], it follows that [itex]|l-m| = |l - f(x_1) + f(x_1) -m| \leq |f(x_1) - l| + |f(x_1) - m| < |l-m|/2 + |l-m|/2 < |l-m|[/itex], which is a contradiction. Therefore, [itex]l=m[/itex].

    Does this work alright? Or is there another way that I should approach this problem? I appreciate any feedback! Thanks!
  2. jcsd
  3. Jul 6, 2010 #2
    Your approach seems to work, but I think it would have been much easier to write out what one of the limits means and then later set x = a + h. This approach also supports the intuitive idea behind this equality. There is also a way to prove this without resorting to the definition of the limit via some manipulations.
  4. Jul 6, 2010 #3
    Looks ok. It is possible to do it more directly. Suppose [tex]\lim_{x\to a}f(x)=L[/tex]. Fix [tex]\epsilon > 0[/tex]. For some [tex]\delta[/tex]-neighbourhood of [tex]a[/tex], [tex]|f(x)-L|<\epsilon[/tex]. Hence, if [tex]|h|<\delta[/tex], [tex]|f(a+h)-L|<\epsilon[/tex] holds, and therefore [tex]\lim_{h\to 0}f(a+h)=L[/tex]. It's essentially the same, though.

    Edit: I'm too slow
    Edit2: Wow, 3 minutes...way too slow
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