# Areas where the limit doesn't exist

I am currently marking for a first year calculus class, and I think my prof made a mistake on the solution sheet for the most recent recitation.

The prof provided a cartesian coordinate system with a function $f(x)$, with vertical asymptotes at $x=-3, x=0$, and a horizantal asymptote at $y=1$.

At the asymptote at $x=0$, the function approaches $+ \infty$ from both sides
(ie $\lim_{x \to 0^+} = \lim_{x \to 0^-} = +\infty$).

Now, one of the questions asks to list all numbers $a$ which $\lim_{x \to a} f(x)$ does not exist. And my prof listed $x=0$ as a point where the limit doesn't exist. But wouldn't the $\lim_{x \to 0} = +\infty$, and thus shouldn't be not listed as a point where the limit doesn't exist?

Thanks for any help

• At the level of an intro calc class, $\lim_{x\to 0} f(x) = +\infty$ is a way of saying that (1) the limit does not exist but (2) it fails to exist in a positive, unbounded kind of way. – Xander Henderson Oct 1 '17 at 1:27
• @XanderHenderson Okay, thanks for clarifying, just wanted to make sure that I was marking correctly – Smeef Oct 1 '17 at 1:31
• As a bit of advice, in the future, when you are grading papers for someone else, it is generally polite to ask that person about how they want things marked before you run to the internet with concerns about the correctness of the solutions provided. It is very possible that your supervisor may have made a mistake, but if that were the case, I'm sure that he or she would be quite happy to have that (politely) pointed out. – Xander Henderson Oct 1 '17 at 1:34
• @XanderHenderson That's good advice, I will do that in the future. I didn't mean to be impolite, it just occured to me while marking – Smeef Oct 1 '17 at 1:37

Typically, we say that $\lim_{x\to a} f(x)$ exists if there is some real number $L$ such that for all $\varepsilon > 0$ we can find some $\delta > 0$ such that $$|x-a| < \delta \implies |f(x) - L| < \varepsilon.$$ When we might write $$\lim_{x\to a} f(x) = +\infty,$$ there is no real number $L$ that does this job. The limit doesn't exist in this case, but the behaviour of the function can be characterized. Specifically, for all $M > 0$, there exists some $\delta > 0$ such that $$|x-a|<\delta \implies f(x) > M.$$ In other words, the function can be made arbitrarily large (and positive) by choosing $x$ close to $a$.