# Limit of the function $f(x) = \begin{cases} 1 & \text{if } x = 1 \\ 0 & \text{otherwise} \end{cases}$

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?

• 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 $1$. – Ulivai Feb 14 at 1:08
• the limit is zero. the fact that x = 1, f(x)=1 doesn't change that. – user29418 Feb 14 at 1:09
• 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. – fleablood Feb 14 at 1:10

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 $$1.$$