Hello,
i suppose the functions are defined on real numbers and have their range also in .
Since composition of cont. functions are still cont., this is right
Regards
Hi,
I'd appreciate some help with this question...
I'm thinking no, since from the definition of limit of a function we require "x not equal to a" for the limit of f(x) to exist. But here, we could have f(x) = l and then g(f(x)) isn't necessarily defined.Suppose that as and as . Must it be true that as ?
Is this right? If not then I could probably find a proof assuming f(x) is not l, but why could I do this?
Many thanks.
We defined continuous functions in 2 ways, f is continuous at a if either "for any sequence zn tending to a, f(zn) tends to a" or
The question doesn't state the range of the functions (the definitions are valid also for complex numbers) and doesn't say they are continuous. I know the composition of 2 continuous functions is continuous, but how do we know we can simply apply this theorem?
The definition of f(x) --> l as x --> a that I had in mind was;
I'm aware also of the similar one using sequences.
It's the additional inclusion of "but not equal to" that's confusing me here. Clearly we can't just ignore it (if we do then the composition DOES tend to k), since f(x) in the question could equal l at some point.
So a counterexample would be eg. f(x) = 3 for all x. Then obviously f(x) --> 3, but the definition of g(y) --> k isn't valid since f(x)-3 is always 0. But this doesn't seem rigorous, so is there a better counterexample, or a proof of why I can "forget" that f(x)=3?