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## Homework Statement

Why does ##\lim_{n \rightarrow \infty} f(x_n) = f(c)## contradict ##\lim_{n \rightarrow \infty} \vert f(x_n) \vert = +\infty##?

edit: where ##c## is in ##[a,b]##

## Homework Equations

Here's the proof I'm reading from Ross page 133.

18.1 Theorem

Let ##f## be a continuous real valued function on a closed interval [a,b]. Then ##f## is a bounded function. Moreover, f assumes its maximum and minimum values on [a,b]; that is there exist ##x_0, y_0## in [a,b] such that ##f(x_0) \le f(x) \le f(y_0)## for all ##x \epsilon [a,b]##.

Proof: Assume ##f## is not bounded on ##[a,b]##. Then to each ##n \epsilon \mathbb{N}## there corresponds an ##x_n \epsilon [a,b]## such that ##\vert f(x_n) > n##. By Bolzano-Weierstrass Theorem 11.5, ##(x_n)## has a subsequence ##(x_{n_k})## that converges to some real number ##x_0## in ##[a,b]##. Since ##f## is continuous at ##x_0##, we have ##\lim_{k \rightarrow \infty} f(x_{n_k}) = f(x_0)##, But we also have ##\lim_{k\rightarrow\infty} \vert f(x_{n_k}) \vert = +\infty##. which is a contradiction. It follows that ##f## is bounded.

//And then another paragraph I haven't really read yet that finishes the proof

## The Attempt at a Solution

1) From def. of limit: For all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##k > N(\varepsilon)## implies ##\vert f(x_{n_k}) - f(x_0) \vert < \varepsilon##. This implies ##\vert f(x_{n_k}) \vert < \vert f(x_0) \vert + \varepsilon##.

2) Since ##f## is not bounded we have for all ##\varepsilon > 0##, there exists some ##k## such that ##\vert f(x_{n_k}) \vert > \vert f(x_0) \vert + \varepsilon##.

So there is only a contradiction between 1) and 2), if the ##k## in 2) is greater than ##N(\varepsilon)## in 1). But how do we guarantee this?

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