2

Suppose $f: D \to \mathbb{R}$ and $x_0$ is a limit point of $D$.

Prove that $\lim_{x\to x_0} = L$ if and only if for every $\varepsilon>0$, there exists a $\delta>0$ such that if $x$ is in $D$ and $0 <|x - x_0|<\delta$ then $|f(x) - L| < \varepsilon$.

I know this proof will have two parts, to prove it true both ways. So the first part goes something like...

Suppose $f$ satisfies the $\varepsilon$-$\delta$ criterion at $x_0$... and going on to prove that $\lim f (x) = L$

The definition I am using for a limit is as follows:

Given a function $f: D \to \mathbb{R}$ and $x_0$ is a limit point of $D$, for a number L, we write $\lim_{x \to x_0} f(x) = L$ provided that whenever ${x_n}$ is a sequence in $D\setminus \{x_0\}$ that converges to $x_0$, $\lim_{n\to \infty} f(x_n) = L $.

Brian Tung
  • 34,160
RSt
  • 21

1 Answers1

2

Suppose for every $\varepsilon>0$, there exists a $\delta>0$ such that if $x$ is in $D$ and $0 <|x - x_0|<\delta$ then $|f(x) - L| < \varepsilon$.

Then take any sequence ${x_n}$ in $D\setminus \{x_0\}$ that converges to $x_0$. Then $\exists N$ such that $\forall n > N$, $0<|x_n-x_0|<\delta$.

So, what can you say about $|f(x_n) - L|$?

Converse: Use contradiction. Suppose $\exists \varepsilon > 0$ such that $\forall \delta >0$ $\exists x \in D$ such that $0 <|x - x_0|<\delta$ but $|f(x) - L| \geq \varepsilon$. So for $\delta_n = \frac{1}{n}$ we will have $x_n \in D$ such that $0 <|x_n - x_0|<\delta$ but $|f(x_n) - L| \geq \varepsilon$.

So we can construct a sequence $(x_n)_{n \in \mathbb{N}}$ in $D\setminus \{x_0\}$ which converges to $x_0$, but for which $f(x_n)$ is always at least $\varepsilon$ away from $L$. We have a contradiction.

Ilham
  • 1,567