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?

  • $\begingroup$ 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 $\lim_{x \to a} f(x) \neq \infty$. $\endgroup$ Oct 19, 2021 at 22:18
  • $\begingroup$ 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? $\endgroup$ Oct 19, 2021 at 23:11
  • $\begingroup$ You may use contradiction and then the triangle inequality to bound $|f(x|$ $\endgroup$ Oct 19, 2021 at 23:36

1 Answer 1


One approach is to argue by contradiction.

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 $$|f(x)-L|<1\text{ for every }x\in\text{dom}[f]\text{ with }0<|x-a|<\delta_1$$ $$\iff L-1<f(x)<L+1\text{ for every }x\in\text{dom}[f]\text{ with }0<|x-a|<\delta_1$$ 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 $$0<|x-a|<\delta_1\text{ and }0<|x-a|<\delta_2$$ $$\implies L-1<f(x)<L+1\text{ and }f(x)>L+1$$ $$\implies f(x)<L+1\text{ and }f(x)>L+1$$ $$\implies L+1<L+1$$ The inequality $L+1<L+1$ 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.


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