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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?

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    $\begingroup$ You elaborate how $\lim f(x)=L$ is defined for real $L$. How is $\lim f(x)=\infty$ defined? $\endgroup$ – Hagen von Eitzen May 8 '20 at 12:58
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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 $x<N$ 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 $x<N$ we have $2+3x^2 >M$ which is what you wanted to prove.

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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, $$\exists L\in\Bbb R:\forall\varepsilon>0,\exists N<0:\bigl(x<N \implies|f(x)-L|<\varepsilon\bigr).$$

To prove this false, we must demonstrate the negation $$\forall L\in\Bbb R,\exists\varepsilon>0:\forall N<0,\neg\bigl(x<N \implies|f(x)-L|<\varepsilon\bigr),$$ or equivalently, $$\forall L\in\Bbb R,\exists\varepsilon>0:\forall N<0,\exists x:(x<N)\wedge \neg\bigl(|f(x)-L|<\varepsilon\bigr).$$

So, we let $L\in\Bbb R$ and $N<0$ be arbitrary, and show that there is some $\varepsilon>0$ and some $x<N$ 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 $x<N$ such that $$2+3x^2\ge L+1?$$

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