# Proving limit at $\infty$ does not exist by definition.

The statement $$\lim_{x\to -\infty}(2+3x^2)=\infty$$

certainly means that the limit doesn't exist at $$-\infty$$. To prove this by the definition of limit requires the use of its negation $$(\vert x-p\vert < \delta) \land\vert f(x)-L\vert\geq\epsilon$$

This function is real valued, so we can't construct any argument by $$p=\infty$$ and so on. Rather, I think it should be sufficient to show that no matter what $$\delta$$ is chosen, when $$p\to \infty$$, the distance $$d(f(x),L)\to \infty$$ and as such is greater than any $$\epsilon>0$$. What steps should I take to make this proof work?

• You elaborate how $\lim f(x)=L$ is defined for real $L$. How is $\lim f(x)=\infty$ defined? – Hagen von Eitzen May 8 '20 at 12:58

In order to prove $$\lim_{x\to -\infty}(2+3x^2)=\infty$$ You need to show that given any $$M>0$$ there exists an $$N<0$$ such that if $$x then $$2+3x^2 >M$$

Given an arbitrary large positive $$M$$, you want $$3x^2>M-2$$ or $$x^2> \frac {M-2}{3}$$

This will be satisfied if $$|x|>\sqrt{ \frac {M-2}{3}}$$ So let $$N=-\sqrt{ \frac {M-2}{3}}$$

Now if $$x we have $$2+3x^2 >M$$ which is what you wanted to prove.

Actually, you're negating the incorrect definition. You're using the definition for the function $$f(x)$$ being continuous at some point $$p$$ of its domain. However, that's not what you're considering. Rather, you're trying to show that there is no value $$L$$ toward which $$f(x)$$ tends as $$x$$ decreases without bound.

If there were such a value $$L\in\Bbb R,$$ that would mean that we could ensure that $$f(x)$$ is as close to $$L$$ as we like, so long as we make $$x$$ sufficiently negative. Symbolically, $$\exists L\in\Bbb R:\forall\varepsilon>0,\exists N<0:\bigl(x

To prove this false, we must demonstrate the negation $$\forall L\in\Bbb R,\exists\varepsilon>0:\forall N<0,\neg\bigl(x or equivalently, $$\forall L\in\Bbb R,\exists\varepsilon>0:\forall N<0,\exists x:(x

So, we let $$L\in\Bbb R$$ and $$N<0$$ be arbitrary, and show that there is some $$\varepsilon>0$$ and some $$x such that $$|f(x)-L|\ge\varepsilon.$$

It actually turns out to work no matter what $$\varepsilon>0$$ we choose, fortunately, so we may as well pick an easy one, like $$\varepsilon=1.$$

Keep in mind that $$|f(x)-L|\ge1$$ is equivalent to $$(f(x)-L\le-1)\vee(f(x)-L\ge1),$$ or $$(f(x)\le L-1)\vee(f(x)\ge L+1).$$ The graph suggests that $$f(x)\ge L+1$$ is what we want to use, here.

So, given our arbitrary $$L\in\Bbb R$$ and $$N<0,$$ can you show that there is some $$x such that $$2+3x^2\ge L+1?$$