Difference between limit tending to infinity and limit doesn't exist In the following question:

I'm getting the answer as LHL=-infinity while RHL= infinity
The answer given to this question is D.
However as far as I know when a limit tends to infinity it does exist. I don't know if I'm wrong, could someone please help.
 A: The formal definition of the existence of a limit at a point $a$:
$$\forall\epsilon>0\exists\delta>0, |x-a|<\delta\implies|f(x)-L|<\epsilon \text{ where } L\in\mathbb{R}$$
So no, the limit must be a constant.
A: Speaking of sequences instead of functions (for which the limit is ultimately defined via sequences), we say that $a$ is the limit of $\{a_n\}$, if for every $\epsilon>0$, at most finitely many $a_k$ are outside the $\epsilon$-neighbourhood of $a$.
$$\tag1 a_n\to a:\iff \forall \epsilon>0\;\exists N\in\Bbb N\;\forall n>N\;|a_n-a|<\epsilon.$$
If an $a$ that is the limit of $a:n$ exists, we say that $\{a_n\}$ is convergent; otherwise it is divergent.
Can $\infty$ be the limit? No, because there is no way to have $|a_n-\infty|<\epsilon$.
However, the concept of a sequence tending to $+\infty$ or $-\infty$ id still of interest. It just doesnt't fit into the pattern $(1)$ and requires a separate definition
$$\tag2 a_n\to+\infty:\iff\forall M\in\Bbb R\; \exists N\in\Bbb N\;\forall n>N\;a_n>M$$
$$\tag3 a_n\to+\infty:\iff\forall M\in\Bbb R\; \exists N\in\Bbb N\;\forall n>N\;a_n<M$$
Thus the notion of "less than arbitrarily small $\epsilon$ away" is replaced with "above arbitrarily large $M$".
By what we previously said, we cannot (or at least should not) say that $a_n$ converges to $\infty$. Instead, in English one usually says that it tends to $\infty$.
A: Indeed, I agree with you, by the usual definition we say that the two side limits exist and we have


*

*$\lim_{x\to1^+}f(x)=+\infty$

*$\lim_{x\to1^-}f(x)=-\infty$


since those are different we say that the limit at $x=1$ doesn't exist and then the function is discontinuos at $x=1$.
