# $\delta$-$\epsilon$ Proof: Prove $\lim_{x\to{a}}f(x)=\infty\implies\lim_{x\to{a}}f(x)$ does not exist, using the formal definition of the limit

The question is describes as follows:

Let $$a \in \mathbb R$$ and let $$f$$ be a function with domain $$\mathbb R$$. Using the formal definition of the limit, prove that: IF $$\lim_{x \to a} f(x) = \infty$$, THEN $$\lim_{x \to a} f(x)$$ diverges.

So far I can only write these two statements as formal mathematical definitions:

$$\lim_{x \to a} f(x) = \infty$$: $$\text{Let }a \in \mathbb R.~ \forall M \in \mathbb R,~ \exists \delta > 0 \text{ such that }0 < |x - a| < \delta \implies f(x) > M$$

$$\lim_{x \to a} f(x)$$ does not exist:

$$\forall L \in \mathbb R,~ \exists \epsilon > 0 \text{ such that } \forall \delta > 0,~ \exists x \in \mathbb R \text{ such that } 0 < |x-a| < \delta \implies |f(x)-L| \ge \epsilon$$

Can anyone please tell me what can I do next?

• Your characterisation of $\lim_{x \to a} f(x)$ not existing is not quite correct: there's no guarantee that $f(x)$ will always lie above the "limit" candidate $L$. Indeed, if you think about it, $\varepsilon$ should make an appearance somewhere other than after the $\exists$ sign. That said, if I were you, I would prove the contrapositive: if $\lim_{x \to a} f(x)$ exists, then $\lim_{x \to a} f(x) \neq \infty$. Oct 19, 2021 at 22:18
• Can you explain (in words, not symbols) how the (random) choice $L = 5$ in the second definition you recite fails due to some valid choice for $M$ in the first definition you recite? Oct 19, 2021 at 23:11
• You may use contradiction and then the triangle inequality to bound $|f(x|$ Oct 19, 2021 at 23:36

Suppose, to the contrary, that $$\lim_{x\to a}f(x)=\infty$$ and $$\lim_{x\to a}f(x)=L$$ for some real number $$L$$. From the definition of $$\lim_{x\to a}f(x)=L$$ with $$\varepsilon=1$$, we can deduce that there is a $$\delta_1>0$$ for which $$|f(x)-L|<1\text{ for every }x\in\text{dom}[f]\text{ with }0<|x-a|<\delta_1$$ $$\iff L-1 Unraveling the definition of $$\lim_{x\to a}f(x)=\infty$$ with $$M=L+1$$, we deduce that there is a $$\delta_2>0$$ for which $$f(x)>L+1\text{ for every }x\in\text{dom}[f]\text{ with }0<|x-a|<\delta_2$$ It follows that for every $$x\in\text{dom}[f]$$ with $$0<|x-a|<\min\{\delta_1,\delta_2\}$$, we have $$0<|x-a|<\delta_1\text{ and }0<|x-a|<\delta_2$$ $$\implies L-1L+1$$ $$\implies f(x)L+1$$ $$\implies L+1 The inequality $$L+1 contradicts the fact that $$L+1=L+1$$. Since our argument is valid, the only possibility is that our original premise/assumption is actually false. Thus, it is impossible to simultaneously have $$\lim_{x\to a}f(x)=\infty$$ and $$\lim_{x\to a}f(x)=L$$ for some real number $$L$$, so if $$\lim_{x\to a}f(x)=\infty$$, it necessarily follows that $$\lim_{x\to a}f(x)=L$$ is false for every $$L\in\mathbb{R}$$, that is, $$\lim_{x\to a}f(x)$$ does not exist.