# $\delta-\varepsilon$ definition of $\lim_{x \to \infty} f(x) = \infty$

I know that

$$\lim_{x \to \infty} f(x)=L \mbox{ means } \forall \varepsilon>0 \:\exists N\:\forall x\:(x>N\rightarrow|f(x)-L|<\varepsilon)$$

$$\lim_{x \to a} f(x)=\infty \mbox{ means } \forall N \:\exists \delta>0\:\forall x(0<|x-a|<\delta|\rightarrow f(x)>N)$$

I've combined these the following way but not sure if it's precise:

$$\lim_{x \to \infty} f(x)=\infty \mbox{ means } \forall M \:\exists N\:\forall x\:(x>N\rightarrow f(x)>M)$$

Is it correct definition of this limit?

• Yes, that is correct. Importantly, the range tolerance is given and then the domain tolerance can be chosen as a function of the range tolerance. – zugzug Mar 31 at 21:48
• It's the $\delta$-$\varepsilon$ limit with neither $\delta$ nor $\varepsilon$ :) ! – Ben Apr 13 at 14:52

For all $$\epsilon \in \mathbb R_+$$ exists $$k$$, such that for all $$x$$
$$|f(x)|>\epsilon,\space \space \text{when} \space x>k.$$