Difference between limit tending to infinity and limit doesn't exist

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In the following question:

enter image description here 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.

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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 $x=1$.

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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.

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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 $\infty$.