0
$\begingroup$

I'm confused about the existence of the limit of a real-valued function $f: \mathbb{R}\rightarrow \mathbb{R}$ and notation.

Suppose that $\forall \epsilon>0$ $\exists \delta_{\epsilon}>0$ s.t. $\forall$ $0<|x-a|<\delta_{\epsilon}$ we have $0<|f(x)-L|<\epsilon$. Then we can write $\lim_{x\rightarrow a}f(x)=L$ which denotes that "the limit of $f(\cdot)$ for $x$ approaching $a$ exists and is equal to $L$".

Suppose instead that the function $f(\cdot)$ goes to $+\infty$ when $x \rightarrow a$. If I have understood correctly the definition of limit, in this case "the limit of $f(\cdot)$ for $x$ approaching $a$ does NOT exists". My question is: why do we write $\lim_{x \rightarrow a} f(x)=+\infty$ when the function has no limit?

$\endgroup$
2
  • $\begingroup$ You are correct. Usually the definition is extended by saying something like $\lim_{x \to a} f(x) = \infty$ iff for all $N$ there is some $\delta>0$ such that if $|x-a| < \delta$ then $f(x) > N$. $\endgroup$
    – copper.hat
    Commented Dec 9, 2015 at 20:24
  • $\begingroup$ Equivalently, one sometimes sees "for each $\epsilon > 0$ there exists $\delta_{\epsilon}>0$ such that $0<\frac{1}{f(x)}<\epsilon$ whenever $0<|x-a|<\delta_{\epsilon}$". $\endgroup$
    – MPW
    Commented Dec 9, 2015 at 20:29

1 Answer 1

Reset to default
1
$\begingroup$

It is just shorthand or notation for a specific case of the limit not existing. The notation contains more information than mere non-existence of a limit though. Indeed if $f(x) \to \infty$ as $x \to a$ then $$ \forall h > 0 : \exists \delta > 0 : x \in \mathbb{R} : |x - a| < \delta : f(x) > h$$

$\endgroup$
2
  • $\begingroup$ Ok, but despite notation, when I'm asked to show that the function has a limit, it means that I have to show that the function has a limit which is $<\infty$ and $> -\infty$, right? $\endgroup$
    – Star
    Commented Dec 9, 2015 at 20:26
  • $\begingroup$ Yes. It is just notation. $\endgroup$
    – Kayle of the Creeks
    Commented Dec 9, 2015 at 21:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .