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!

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# Limit Proof

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