# How to express the following limit in epsilon-delta language

1. Nov 23, 2006

### Jacobpm64

Express the following as an epsilon-delta proof (to show that it is continuous):

$$\lim_{x\rightarrow - \infty}f(x) = L$$

Can I get some ideas on this one?

2. Nov 24, 2006

### matt grime

No one can do that question: you have failed to say what f is.

3. Nov 24, 2006

### HallsofIvy

Staff Emeritus
Just express that in terms of $\epsilon$, $\delta$ for general f and L? Then just use the definition of limit at infinity: the "standard" definition of limit, at some number a, is "Given any $\epsilon> 0$, there exist $\delta$ such that if $|x- a|< \delta$ then $|f(x)- L|< \epsilon$." If you really mean "at infinity" then "$|x- a|< \delta$" has to become "x sufficiently large" or "x> X" for some number X.

The definition, then, is
Given any $\epsilon> 0$, there exist a real number X such if x> X then $|f(x)- L|< \epsilon$.

The reason I said 'If you really mean "at infinity"' is that you also say "to show that it is continuous". You are taking the limit as x goes to infinity and a function is never continuous "at infinity". Taking the limit as x GOES to infinity simply means "as x gets larger without bound". "Infinity" is not a specific number and f is not defined "at infinity".
I wonder if you don't actually mean "limit as x goes to a" for some number a.