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When I first studied limits I have been taught that the limits tells us that the function will gets closer to some value when the domain of the function gets closer to some value.

then I studied the (ε-δ) definition and the (ε-δ) definition makes perfect sense with what I sudied before.

if you truly approaching a value then you are capable to make the function within a tolerance (ε) from the value you are approaching.

But this defintion works very well to me except in the constant function ,in the constant function the function stays the same all the time when x get closer and closer to some value the function doesn't gets closer and closer to any value,the function stays the same all the time so how we define the limit of a cosntant function?

Rodrigo de Azevedo
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Mans
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    The limit of a constant value is that constant. It is always close to it. – Carla only proves trivial prop Feb 08 '22 at 16:36
  • The definition is exactly the same, and you can prove that a constant function has itself as it's limit. – Lorago Feb 08 '22 at 16:42
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    Thinking of limits in the "closer and closer" sense is good for building (imperfect) intuition. But ultimately, everything in mathematics has a rigorous definition, and that definition determines the truth or falsity of statements; when our intuition is in conflict with the consequences of the definitions, we simply have to alter our intuition (or accept that our intuition is incorrect here). – Greg Martin Feb 08 '22 at 17:25

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Let $D\subset\Bbb R$, let $a\in\Bbb R$ such that every interval $(a-\varepsilon,a+\varepsilon)$ contains some point of $D$, let $f\colon D\longrightarrow\Bbb R$ be a function, and let $l\in\Bbb R$. We say that the limit of $f$ at $a$ is $l$ if$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in D):|x-a|<\delta\implies\bigl|f(x)-l\bigr|<\varepsilon.$$If it turns out that $f$ is constant, that means that, for some $k\in\Bbb R$, $(\forall x\in D):f(x)=k$. But then the limit of $f$ at $a$ is $k$. In fact, if $\varepsilon>0$, then you can take any $\delta>0$, and then, if $x\in D$,$$|x-a|<\delta\implies\bigl|f(x)-k\bigr|=0<\varepsilon.$$

Concerning your informal approach at the end of your post, what happens is this: as $x$ gets closer and closer to $a$, $f(x)$ is already at $k$.

José Carlos Santos
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    But it's already at (K) the function doesn't even appraoach any thing it's just stays the same I understand the proof of the epislon delta definition but I want to know how the intuition of limits in the constant function works – Mans Feb 08 '22 at 17:03
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    Already being at a spot is an extreme case of approaching that spot. – José Carlos Santos Feb 08 '22 at 17:04
  • Thank you for your answering – Mans Feb 08 '22 at 17:08