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If someone proposes the problem:

Calculate the limit: $\lim_{x\rightarrow 2}\frac{x-2}{2-\sqrt{4}}$

For me the limit does not exist because in fact the function $\frac{x-2}{2-\sqrt{4}}$ does not exist. However, it is also true that the problem is misspelled. But my question is that if they already pose the problem to you like this, what is the correct thing to say, you cannot write such a limit (misspelled) or it does not exist.

SSS1
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  • Because $2-\sqrt{4}=2-2=0$. – SSS1 Sep 08 '21 at 21:41
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    The notation for $\lim\limits_{x \to 2} f(x)$ only makes sense if $f$ is defined over $(2 - \epsilon, 2) \cup (2, 2 + \epsilon)$ for some positive $\epsilon$. So I would say that the question is ill-formed. – Mark Saving Sep 08 '21 at 21:43
  • Yep! In my head, I was just treating $\sqrt{4}$ as a symbol. – march Sep 08 '21 at 21:44

1 Answers1

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I would say that the expression is not well defined because it is equivalent to $\frac{x-2}0$ and therefore it is meaningless consider the limit at $x=2$ which is not a cluster point.

We could define a value $L$ for the function at $x=2$ and then extend the notion of limit to isolate points such that

$$\lim_{x\rightarrow 2}\frac{x-2}{2-\sqrt{4}}=L$$

as discussed in the following references

user
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  • Cluster point of what? – Rob Arthan Sep 08 '21 at 21:57
  • @RobArthan I mean a cluster point of the domain. We could also define the function at $x=2$ but also in this case the limit problem would be not well posed. I edit to make it more clear. – user Sep 08 '21 at 22:01