- #1

- 1,462

- 44

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

- #2

- 941

- 394

I may be able to find a specific statement of this.

- #3

- 941

- 394

Suppose there is a [itex] \delta > 0[/itex] such that [itex]f(x) = g(x)[/itex] when [itex] 0 < |x - a| < \delta[/itex]. Prove that [itex] \lim_{x \to a} f(x) = \lim_{x \to a} g(x)[/itex]. In other words, the limit only depends on the values of [itex]f(x)[/itex] for [itex]x[/itex] near [itex]a[/itex].

If you're familiar with epsilon-delta proofs (the limit approaches [itex]L[/itex] if [itex]0 < |x - a| < \delta [/itex] implies [itex] |f(x) - L| < \epsilon[/itex]), then this is easy to see: We know that [itex]f(x) = g(x)[/itex] for some neighborhood around [itex]a[/itex], that is, [itex]f(x) = g(x)[/itex] when [itex]|x - a| < \delta '[/itex]. Furthermore, if [itex]\lim_{x \to a} f(x) = L[/itex] then given an [itex]\epsilon > 0[/itex], then there exists some [itex]\delta > 0[/itex] such that [itex]0 < |x - a| < \delta \implies |f(x) - L| < \epsilon[/itex]. Now just make sure [itex]\delta ' < \delta[/itex], and then we know [itex]g(x)[/itex] is equivalent to [itex]f(x)[/itex], so we can replace it in our mathematical statement: [itex] 0 < |x - a| < \delta ' \implies |g(x) - L| < \epsilon[/itex]. Thus [itex]f[/itex] and [itex]g[/itex] both approach [itex]L[/itex] as [itex]x \to a[/itex].

Intuitively, this just means that we only care about the neighborhoods

Share: