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Consider function $f(x) = \begin{cases} 1 & \text{if } x = 1 \\ 0 & \text{otherwise} \end{cases} $

I am trying to find limit of this function when $x \to 1$. By two-sided limit theorem, it can concluded $$\lim_{x \to 1^{-}}f(x) = \lim_{x \to1^{+}}f(x) = 0 \implies \lim_{x \to 1}f(x) = 0.$$

But when trying to apply the definition of the limit there is a problem: we want to bound $|f(x) - 0| = |f(x)|$ by $\epsilon$, $|x - 1| < \delta$ and when $x = 1$ we have $|f(x)| = 1$ which is not bounded for any $\epsilon > 0$. The question is: does the limit even exist?

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    $\begingroup$ the limit goes to zero, but i don't understand why did you consider the case when $x=1$ to evaluate the limit on trying to apply the definition of the limit, you should evaluate values of $x$ close to $1$ but not actually on $1$. $\endgroup$ – Ulivai Feb 14 at 1:08
  • $\begingroup$ the limit is zero. the fact that x = 1, f(x)=1 doesn't change that. $\endgroup$ – user29418 Feb 14 at 1:09
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    $\begingroup$ When selecting an $x$ so that $0<|x-a| < \delta$ you only select $x \ne a$. So you are not allowed to select $x = 1$ you must select an $x \ne 1$. That is you must have $0 < |x-1| < \delta$. If $x =1$ you have $0=|x-1|$ not.... well, not "not allowed" but ... not relevant. $\endgroup$ – fleablood Feb 14 at 1:10
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Where you wrote $|x-1|<\delta,$ you need $0<|x-1|<\delta.$

I.e. you can make $f(x)$ as close to the limit as you want by making $x$ close enough, but not equal, to $1.$

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  • $\begingroup$ Thank you, yes indeed it answers the question! Completely forgot about it. $\endgroup$ – stackoverload Feb 14 at 1:10

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