Question

Can anyone please give an example of why the following definition of lim f(x) = L (as x -> a) is NOT correct ?:

For every real δ>0 there exists a real ε>0 such that for all real x, 0<|x-a|<δ then |f(x)-L|<ε.

I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work.

Answer 1

the definition is
\[ \large \lim_{x\to a}f(x)=L\quad\Leftrightarrow \]
\[ \large (\forall\varepsilon>0)(\exists\delta_\varepsilon>0)(\forall x)(0<|x-a|<
\delta_\varepsilon\to|f(x)-L|<\varepsilon) \]

Answer 2

helder_edwin: thanks, but that is the definition of limit, I already know that. What I'm trying to know is why the alternative one in my original post doesn't work by giving an example in which it doesn't work.

Answer 3

oh. OK

Answer 4

there will always exist such an epsilon whatever the limit as you can make epsilon as large as you want it to be regardless of the limit so this 'definition' isn't really saying anything at all

Answer 5

Thanks, that's it!

Answer 6

sorry. i was helping someone else.
i was thinking of a function like this
\[ \large f(x)=\begin{cases}
0,\quad &\hbox{if \(x<\delta\)}\\
\delta/2,\quad&\hbox{if \(x\geq\delta\)}
\end{cases} \]

Answer 7

the "definition" u posted makes the limit at \(x=\delta\) exist !!!