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(This question: Limits of functions at end points of its domain is same as mine but there is no verified answer and had left me more confused. I'll be very thankful if complete answer is provided) Graph of the function Consider this graph of a function $y=f(x)$ $x ∈ (0,4)∪(5, \infty)$ . I want to know that what is the value of $\lim f(x)$ as :

$(a)$ $x$ tends to $0$ ( Is it $-\infty$ or limit don't exist ? )

$(b)$ $x$ tends to $5$ ( Is it $2$ or limit don't exist?)

I am having confusion in it from many days. We can't calculate limit from left side in both cases. However, a value approaches if we move from right side only, and in one case it's negative infinity and in one case it's a finite number.

Please clarify it and also tell that how to solve these types of limits if they are provided to us in equations instead of graphical form.

N. F. Taussig
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An_Elephant
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    When we say a limit is infinite, technically the limit doesn't exist. – Ivan Kaznacheyeu Sep 16 '22 at 10:24
  • @IvanKaznacheyeu Thanks very much . It mean answer of ($a$) is limit don't exist ? And what about ($b$) ? Please expand your comment to a complete answer. – An_Elephant Sep 16 '22 at 10:32
  • As for second question: you cannot apply definition of limit of function defined in interval containing the point (maybe except the point) to this case. Then you need to use some other definition. There is natural generalization: $\lim_{x\to a} f(x)=A$ if for every $\varepsilon > 0$ exists $\delta > 0$ such that: 1) exists at least one $x$ in domain of $f$ such that $0<|x-a|<\delta$; 2) for every $x$ in domain of $f$ satisfying $0<|x-a|<\delta$ inequality $|f(x)-A|<\varepsilon$ is also satisfied. – Ivan Kaznacheyeu Sep 16 '22 at 10:38
  • I'm not sure about case when $x=a$ is standalone point of domain. – Ivan Kaznacheyeu Sep 16 '22 at 10:40
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    "When we say a limit is infinite, technically the limit doesn't exist": not quite. "Obviously it depends on the definition of "exists".". – Anne Bauval Sep 16 '22 at 10:40
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    See this question instead: https://math.stackexchange.com/questions/637280/limit-of-sqrt-x-as-x-approaches-0 – Hans Lundmark Sep 16 '22 at 10:59
  • @HansLundmark Thanks. This helped and clearly answer ($b$) .From the kind comments, I conclude that at end points or where the function breaks, the limit is said to be exist if the approaching number is finite . And it don't exist if the limit approaches infinity at the end point. Thanks everyone – An_Elephant Sep 16 '22 at 13:50
  • The reason why we say a limit of $\pm\infty$ doesn't exist is because we are taking the limit within the set of real numbers. Neither $+\infty$ nor $-\infty$ is a real number. So while we have a concept of what it means for the limit to go to each, they still do not exist in the real numbers. However, we are free to define whatever is most suitable to our needs. We can grab a couple handy non-real-number objects, label them "$-\infty$" and "$+\infty$" and toss them in with the real numbers to form the extended reals $\overline{\Bbb R}$, add some axioms to handle them and there you go. – Paul Sinclair Sep 17 '22 at 12:55
  • However the extended reals are not as nicely behaved as the real numbers, and when working with them we end up with a lot of special cases. This is why we usually prefer to treat $\pm\infty$ as a shorthand means to describe particular ways in which a limit can diverge, instead of as actual objects to which the limit can converge. – Paul Sinclair Sep 17 '22 at 13:01
  • Thanks Paul. But why we have two different conventions of real numbers? – An_Elephant Sep 17 '22 at 14:02
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    @An_Elephant just like we extend naturals to integers to suit our needs (eg, solving x+2=0), similarly we extend the real numbers to either complex numbers or hyperreals to suit our needs. – insipidintegrator Sep 17 '22 at 15:13

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