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So recently I was studying delta epsilon method to prove limit. And I saw some easy limits to be proved formally using this method.

So I tried to frame some question of my own and I came across this limit

$$0(\displaystyle{\lim_{x \to \infty}} x)$$

My attempt:

Let x=$\frac 1 t$ then limit translates to :

$$0(\displaystyle{\lim_{t \to 0}} \frac 1 t)$$

which can be written as(or can it I have no idea) $$(\displaystyle{\lim_{t \to 0}} \frac 0 t)$$

After this I have no idea I cannot figure what is f(x)?(is it 0?? Doesn't seem right) or how to start with $ \left\lvert x-0 \right\rvert \lt \delta$ and arrive at a conclusion.

The limits seems to tend to zero graphically and intuitively to me but I cannot formalise.

Question:

How can we prove limit $$0(\displaystyle{\lim_{x \to \infty}} x)$$ exists or does not exists ?

(I am sorry if this a silly question to begin with)

Hrishabh Nayal
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    I'd say $\lim_{x\to\infty}x$ was not meaningful. – Angina Seng Apr 01 '20 at 15:11
  • It is the first time that I see such an symbol! – dmtri Apr 01 '20 at 15:11
  • @Angina Seng so the question is stupid after all (thought so) but I can't help imagine can the question be translated to $(\displaystyle{\lim_{t \to 0}} \frac 0 t)$ is that meaningful? – Hrishabh Nayal Apr 01 '20 at 15:19
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    @HrishabhNayal You can't move the $0$ into the limit since the original limit without the constant does not exist. If it were something like $0\cdot \lim\limits_{t\to \color{red}{1}}\frac1t$, it would be fine to rearrange it to $\lim\limits_{t\to \color{red}{1}}\frac0t$ since in that case the limit would have existed. – Jam Apr 01 '20 at 15:22

1 Answers1

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The expression is undefined since $\lim\limits_{x\to\infty}x$ (which is computed before $0$ is introduced) does not exist, or is $+\infty$. See (mathcentre: Limits of functions pp. 3-4) and (Question 2347692). Your rearrangement using $a\lim\limits_{x\to c}f(x)=\lim\limits_{x\to c}af(x)$ does not hold because it is only valid when the limit in the LHS exists. However, you would be correct that $\lim\limits_{x\to \infty}0\cdot x=0$.

Jam
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  • Wait I do not get why $a\lim\limits_{x\to c}f(x)=\lim\limits_{x\to c}af(x)$ holds only when LHS exists? – Hrishabh Nayal Apr 01 '20 at 15:24
  • @HrishabhNayal Think of it like an algorithm where each operation is taken in a certain order. On the LHS, you (1) start with the function (2) take the limit and (3) multiply it by the constant. On the RHS, you swap steps (2) and (3). If we get stuck at step (2) for the LHS, then the whole LHS is undefined and we can't use it in an equation. It's like how an equation like $y=\frac1x$ would only hold when $x\ne 0$ since both sides need to exist. Does that make sense? – Jam Apr 01 '20 at 15:30
  • Another way of thinking about your problem is asking whether $0\cdot (1+1+1+\ldots)=0+0+0+\ldots$ is true. The question of whether $0$ or $(1+1+\ldots) = \infty$ is the more dominant term is impossible to answer, so we call the LHS an indeterminate form. – Jam Apr 01 '20 at 15:37