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:(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Limit Proof

**Physics Forums | Science Articles, Homework Help, Discussion**