1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ – zugzug Mar 31 at 21:48
  • $\begingroup$ It's the $\delta$-$\varepsilon$ limit with neither $\delta$ nor $\varepsilon$ :) ! $\endgroup$ – Ben Apr 13 at 14:52
0
$\begingroup$

Definition:

For all $\epsilon \in \mathbb R_+$ exists $k$, such that for all $x$

$$|f(x)|>\epsilon,\space \space \text{when} \space x>k.$$

So yes it appears correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.