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I have a difficulty in proving this statement:

A function is continuous if always $\lim_{n\to\infty}a_n=a$ implies $$\lim_{n\to\infty}f(a_n)=f(\lim_{n\to\infty}a_n),$$

Could anyone help me please?

Emptymind
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You know that if a function is continuous, there is some $\delta$ such that for any $\epsilon$, $|x-y|<\delta$ $\implies |f(x)-f(y)|<\epsilon$. What can you say about the fact that a sequence $(a_n)_n$ converges to $a$? Try to formulate this in terms of $\delta$'s, then the result will follow. (Note that $f(\lim_{n\to \infty}a_n) = f(a)$).

rednexela1941
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    This is backwards in several ways. (1) When you're trying to prove that the function is continuous, it doesn't help to state that such-and-such would hold if it is. (2) Unless the function is constant, there certainly is no $\delta$ that will work for every $\varepsilon$ like you claim. (3) Without explicitly constructing a sequence $(a_n)n$ you have nothing to say about it; the premise of the statement merely promises that _if you somehow find a sequence that converges, then the limit of the function value is the function value at the limit. – hmakholm left over Monica Oct 22 '17 at 12:32
  • Reread the question, we are understanding it as two different things. I took this as the statement that: if f is continuous (assumed to be on some set) then if the sequence property listed above holds, (which is true and in fact is if and only if as this is an equivalent definition of continuity). (1) - we are not trying to show it is continuous, we are assuming it. (2) - not what I claimed, i said for each $x$ and $\epsilon$, there is a $\delta$ (3) - no need to construct the sequence, we are given that the sequence converges, which is the only relevant property. Cheers. – rednexela1941 Oct 26 '17 at 01:24