- #1

Yagoda

- 46

- 0

## Homework Statement

The function f is defined on a neighborhood N of [itex]\bar{x}[/itex]. Show that

[itex] \lim_{x \rightarrow \bar{x}} f(x) = L [/itex] if and only if [itex] \lim_{n \rightarrow \infty} f(x_n) = L [/itex] when [itex] \{x-n\}[/itex] is a sequence of points in N with [itex]\lim_{n \rightarrow \infty} x_n = \bar{x} [/itex].

## Homework Equations

## The Attempt at a Solution

I think I have the necessity, but I am having trouble tightening up the other direction. What I have is that if [itex]\lim_n \rightarrow \infty x_n = \bar{x} [/itex] then [itex]\forall \delta > 0, \exists N_1 : |x_n - \bar{x}| < \delta \text{ when } n > N_1[/itex] and that [itex]\forall \epsilon > 0, \exists N_2 : |f(x_n) - L| < \epsilon \text{ when } n > N_2 [/itex].

Let [itex] N = \max\{N_1,N_2\} [/itex] there is an N so that if [itex]|x_n - \bar{x}| < \delta [/itex] then [itex]|f(x_n) - L| < \epsilon[/itex] for any [itex]\epsilon < 0[/itex]. Which is close to the definition of the limit of f, but how can I generalize it to [itex]|f(x) - L| < \epsilon[/itex] rather than [itex]|f(x_n) - L| < \epsilon[/itex] or is that not an issue?