# limit of a function at infinity from the definition perspective

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.