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Considering the limit

$$\lim\limits_{x \to 3} \frac{2x}{x-3}$$

The left and right approach to this will end up with some thing like :

$$\lim\limits_{x \to 3^+} \frac{2x}{x-3} = \infty$$

$$\lim\limits_{x \to 3^-} \frac{2x}{x-3} = -\infty$$

The book that I'm use says that this just anther way to represent limit still not exist but because the arbitrary increasing and decreasing in the function values as we tend to 3 we use ∞ to represent this case.

Can I conclude The Following

$$\lim\limits_{x \to 3} \frac{2x}{x-3} = \lim\limits_{x \to 3^+} \frac{2x}{x-3} = \lim\limits_{x \to 3^-} \frac{2x}{x-3} = \text{D.N.E}$$

based on all limits actually are not exist

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  • $\begingroup$ The LHS and RHS limits are clearly not equal ($-\infty$ and $+\infty$). $\endgroup$ – KM101 Jan 6 '19 at 19:52
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There are some areas (such as computability theory) where it is a useful convention to say that $$ \mathit{expr}_1 = \mathit{expr}_2 $$ means, "either both expressions are defined with the same value or neither of the expressions are defined".

But basic real analysis is not one of those areas. (And even in the areas where the convention is used, it is good form to explicitly say that you're using it before jumping right into equations).

The usual convention for $=$ is more or less that at best $\mathit{expr}_1 = \mathit{expr}_2$ is false when one or both of the expressions is undefined. At worst the expression is considered to be nonsense if we write something that depends on its truth value in a context where we're not sure both are defined.


Muddying the waters a bit further we have the notation $\lim_{x\to a}f(x)=+\infty$. The most common way to define this notation is that the entire combination of ink shapes "$\lim\cdots=+\infty$" is a single symbol and the result is not actually an equation where "$=$" has its usual meaning. The whole thing is just a conventional way to claim $f(x)$ fails in a particular way to have a finite limit.

However, it is also possible to declare that you're working in the extended real line, in which case $+\infty$ and $-\infty$ are honest points in your topological space that are fully capable of being values of a limit expression.

Depending on which of these conventions you use, the claim, for example $$ \lim_{x\to 2} \frac{1}{(x-2)^2} = \lim_{x\to \pi/2}\tan^2 x $$ could either be considered to be perfectly good (and true), or to be nonsense. Since you don't know which definitions your readers will favor, it is generally good form not to write something like this unless you also point out that you're considering limits in the extended real line.


If you want to consider both limits of the form $\lim f(x)=+\infty$ and of the form $\lim f(x)=\pm\infty$ (where the latter form is satisfied by, for example, $\lim_{x\to 0}1/x = \pm\infty$), then even the extended real line will not save you, because a single function can now have more than one true limit. Then you have to fall back to the "naive" way out, saying that $\lim f(x)=\cdots$ is not actually an equation, but an indivisible claim about the behavior of $f$, and that "$\lim f(x)$" is not to be understood as a (single-valued) expression.

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  • $\begingroup$ To the last section: it might be worth noting that there is always the projective real line if you just want a single $\infty$ which doesn't care which way you went to get there. (Admittedly, I can't think of a single instance in real analysis where this is useful, but it does express a sensible claim of "the function gets really far from $0$ here") $\endgroup$ – Milo Brandt Jan 6 '19 at 21:13
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    $\begingroup$ @MiloBrandt: True enough: you can get either one-infinity limits or two-infinity limits by choosing the superspace appopriately. The point I was trying to make was that you don't get to mix both concepts and still act like there's a "the" limit. $\endgroup$ – hmakholm left over Monica Jan 6 '19 at 21:15
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Definition: a limit of a function in a point exists if right and left limits exist and are equal. Then the equal value is called the limit of the function in that point.

For example if $$\lim_{x\to 0^+}f(x)=1\\\lim_{x\to 0^-}f(x)=-3$$then $$\lim_{x\to 0}f(x)=\text{D.N.E.}$$furthermore if $$\lim_{x\to 3^+}f(x)=\infty\\\lim_{x\to 3^-}f(x)=-\infty$$(which is the case of this problem) we obtain $$\lim_{x\to 3^+}f(x)\ne \lim_{x\to 3^-}f(x)$$and consequently $$\lim_{x\to 3}f(x)=\text{D.N.E.}$$

also take a look at the following graph

https://www.desmos.com/calculator/xatgmwuvsp

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No,you can't write like that.but you should write, $$\lim\limits_{x \to 3^+} \frac{1}{x-3}\ne \lim\limits_{x \to 3^-} \frac{1}{x-3}$$ $$\lim\limits_{x \to 3} \frac{2x}{x-3}=D.N.E $$

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