# How are the epsilon delta definitions equivalent?

The epsilon–delta limit definition $$1$$:

A function $$f(x)$$ from $$\mathbf{R}$$ to $$\mathbf{R}$$ has limit $$L$$ at point $$x_0 \in \mathbf{R}$$ if: $$\bbox[yellow] {\text{for every \epsilon > 0 there exists a \delta > 0 such that whenever |x-x_0| < \delta then}\ |f(x)-L| < \epsilon}$$

Intuitive definition $$2$$:

A function $$f(x)$$ from $$\mathbf{R}$$ to $$\mathbf{R}$$ has limit $$L$$ at point $$x_0 \in \mathbf{R}$$ if: $$\bbox[yellow] {\text{I can get f(x) as close as I want to L by choosing x close enough to x_0}}$$

Can anybody explain in a step-by-step manner how the definition $$1$$ implies definition $$2$$?

• The intuitive definition should be: I can get $f(x)$ as close as a want to $L$ as long as I choose $x$ close enough to $x_0$. – Tychonoff3000 Jun 13 at 13:06
• The intuitive "definition" is wrong and does not follow from the first definition. – StackTD Jun 13 at 13:10
• Ok I have edited – Joe Jun 13 at 13:26

## 1 Answer

Unfortunately, both definitions are incorrect.

The first definition says, rather, that $$f$$ is continuous at $$x_0$$ and $$f(x_0)=L.$$ Instead, it should read

if for all $$\epsilon>0,$$ there exists $$\delta>0$$ such that whenever $$0<\left\lvert x-x_0\right\rvert<\delta$$ then $$\left\lvert f(x)-L\right\rvert<\epsilon.$$

Consider, for example, the function $$f(x)=\begin{cases}1 & x=0\\0 & x\ne 0.\end{cases}$$

Readily, $$f$$ has limit $$0$$ at $$x_0=0,$$ but doesn't satisfy the given definition, since for $$0<\epsilon<1,$$ no appropriate $$\delta$$ can be found.

Moreover, the second definition is inaccurate, as well, as any constant function $$f(x)=L$$ demonstrates. There is no way for $$f(x)$$ to get closer to $$L$$ as $$x$$ gets closer to $$x_0,$$ since $$f(x)$$ is always equal to $$f(x_0)$$!

Instead, I tend to think about it this way. Saying that $$f$$ has limit $$L$$ at $$x_0$$ says that we're guaranteed to be able to get $$f(x)$$ as close to $$L$$ as we like (that is, within $$\epsilon$$ for any positive $$\epsilon$$ we choose), so long as we get $$x$$ sufficiently close to (within some positive $$\delta$$ of), but not equal to, $$x_0.$$

• I am asking if the definition is satisfied, how would it imply that we can get $f(x)$ as close as we want to $L$ by choosing $x$ close enough to $x_0$? Please explain the inner reasonings step by step. – Joe Jun 13 at 13:39
• It doesn't imply that. It simply states it formally. If we want $f(x)$ to be within $\epsilon$ of $L$--that is, if we want $|f(x)-L|<\epsilon$--then we're guaranteed that there is a positive $\delta$ such that for $x$ within $\delta$ of (but not equal to) $x_0$--that is, for $x$ such that $0<|x-x_0|<\delta$--we get what we want. – Cameron Buie Jun 13 at 13:44