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Suppose we have a sequence $\{x_n\}$ which converges to $x$ (which means that there exists an $N$ such that ....) and we know that $\lim_{n->\infty}f(x_n)$ exists (which means that there exists an $M$ such that ....). Is there any relation between $M$ and $N$ ?

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It depends on the function. Of course, for specific functions you could make statements. Like for $f(x) = x$ you could use $M=N$. – Brusko651 Sep 10 '13 at 13:51
It depends on the sequence as well. With the possibility that your sequence doesn't converge in an "even" manner (like, it closes in fast, then stops up and hovers a bit some short distance away, then go back in closer to $x$, then a little bit out again and so on) then you really can't say anything without knowing the function and sequence in particular. – Arthur Sep 10 '13 at 13:58
up vote 1 down vote accepted

The definition of the limit is actually "for all $\epsilon$ there exists $N$..." so the $N$ and $M$ depends on the choice of $\epsilon$.

Now you can ask is for a fixed $\epsilon$ there is a relationship between $N$ and $M$. This depends on the function: it can "distort" arbitrarily $N$ and $M$, so in general we cannot say anything.

For instance take $f(x)=x^{100}$, $\epsilon=1/100$ and $x_n=1/n$. You get $N=100$ and $M=2$.

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