# Epsilon Delta Limit Proofs at and going to infinity.

So I understand the concept of epsilon delta limit proofs with linear functions, easy enough, and I am still shaky about doing it with non linear but I am slowly understanding that. I don't quite understand how to tackle them with you have infinity involved. My professor uses M's and N's and I really don't know what these are supposed to represent in terms of the technical definition we are using here.

One of the problems I have to look at is: $\lim_{x \to \infty} e^x = \infty$

Can anyone give me some other similar examples, not necessarily with $f(x) =e^x$, but any of these proofs that involve infinity, because the normal definition no longer works as written and I don't really know where to begin.

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Here, you need to prove that $e^x$ is unbounded.

Given any $\epsilon\gt 0, \exists x_0\gt 0$ such that $|f(x)|\gt \epsilon$ $\forall x\gt x_0$.

Choose $\epsilon\gt 0$. Now, we need to find corresponding $x_0$ such that $|e^x|\gt \epsilon$ $\forall x\gt x_0$.

If we take $x_0=\ln\epsilon$, then $|e^x|\gt|e^{x_0}|(=\epsilon)\implies |e^x|\gt \epsilon$ $\forall x\gt \ln\epsilon$, Hence $e^x$ is unbounded and positive and thus $\lim_{x\to\infty}e^x=\infty$

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If you want to show something like $$\lim_{x\to\infty} x^2=\infty,$$ then the statement is that, if you go to large enough values of $x$, then the function gets arbitrarily large. More formally, given an arbitrary $M>0$, we can produce an $x_0$ such that, for all $x>x_0$, we will have $x^2>M$. This is what it means to have an infinite limit. In the case of $x^2$, simply take $x_0=\sqrt{M}$.

Similarly, with something like $$\lim_{x\to 0}\frac{1}{x^2}=\infty,$$ the statement is that if you get close enough to $x=0$, the function gets arbitrarily large. In this case, given $M>0$, we have that $|x|<\frac{1}{\sqrt{M}}$ implies $\frac{1}{x^2}>M$.

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