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By definition from Wiki, for $f(x)$ a real function, the limit of $f$ as $x$ approaches infinity is $L$ if for all $\epsilon > 0$, there exists a $M$ such that $|f(x) - L| < \epsilon$ whenever $x > M$.
Now I can prove a function $g(x)$ that for all $\epsilon > 0$, there exists a $M$ such that $|g(x) - L| < 2\epsilon$ whenever $x > M$. Can I say $\underset{x \rightarrow \infty}{\lim}~g(x) = L$?
Yes. Let $\epsilon'=2\epsilon$, and suppose $\epsilon'>0$. Clearly, $\epsilon>0$, so the condition implies there exists an $M$ so that $|g(x)-L|<2\epsilon$ whenever $x>M$. But this is the same thing as saying that given any $\epsilon'>0$, there exists an $M$ so that $|g(x)-L|<\epsilon'$ for any $x>M$, which is exactly the original definition.
Yes, of course. Simply take different letters in both sentences, for example $\epsilon_1$ in first and $\epsilon_2$ in second, and prove, that they are equivalent: suppose 1-st is true for $\forall \epsilon_1$. Then take $\epsilon_1 = 2 \frac{\epsilon_1}{2}$ and because 1-st is true also for $\frac{\epsilon_1}{2}$, then you have second form. Same in reverse.
