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Sorry if my title doesn't explain very well. (I don't know how should I translate it). But I have this problem:
I had this function: $$ f(x) = \begin{cases} 1 & \text{$x$ $\neq$ 0} \\ 0 & \text{$x$ = 0} \end{cases} $$
Can anyone help me why $$\lim_{x\to0} f(x) = 1$$ and not 0?

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  • $\begingroup$ The name is "piecewise (defined) function", if I'm not mistaken. $\endgroup$
    – user562983
    Commented Jun 16, 2018 at 21:33

2 Answers 2

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$$\lim_{x\to0,x\in\mathbb R^*}f (x)=1$$ because $$(\forall x\in\mathbb R^*)\;\;f (x)=1$$ and we write $$\lim_{x\to0,x\ne0}f (x)=1$$ but $$\lim_{x\to0,x\in\mathbb R}f (x) $$ doesn't exist.

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Simply because, when calculating the limit of a function as $x$ tends to a certain value $V$, the point where $x=V$ is not included. Since $f(x)=1$ everywhere except $x=0$, and here $V=0$, this just means that the limit must be $1$.

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  • $\begingroup$ But what about 0 if $x = 0$ condition? It confuses me. The function is 1 if $x$ if not 0 but it is 0 if $x = 0$. $\endgroup$ Commented Jun 16, 2018 at 21:33
  • $\begingroup$ Your function says the opposite. Your function says it is $0$ if $x=0$ and $1$ otherwise, and the $\lim_{x\to0}$ only uses $x\ne0$ $\endgroup$ Commented Jun 16, 2018 at 21:36
  • $\begingroup$ I mistyped earlier in my comment and I modify it now. So, (0 if $x = 0$) doesn't make any changes (should not be considered) in the limits of the function? $\endgroup$ Commented Jun 16, 2018 at 21:38
  • $\begingroup$ Since we are talking about $x$ approaching $0$, we are talking about what happens before $x=0$ not when $x=0$ $\endgroup$ Commented Jun 16, 2018 at 22:57

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