How to express the following limit in epsilon-delta language

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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?
 
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No one can do that question: you have failed to say what f is.
 
Jacobpm64 said:
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?
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.
 
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