I want to examine
$$\lim_{x \to a} \left\lfloor x\right\rfloor $$ using the $\epsilon - \delta$ definition.
I examine two cases where $a$ is an integer and then is not an integer. When $a$ is an integer this is simple because I do the left and right hand limits where in both cases $|f(x)-l|=0$ so...
Away from integers, $f(x)=\lfloor x\rfloor$ is locally constant. In the vicinity of an integer, the function is constant on each side of the integer, but the constants on each side are different.
"I do the left and right hand limits where in both cases $|f(x)−l|=0$" are you sure about that? What does it happen to the left limit when $a$ is an integer?
@Marco I know that? $a$ is not equal to $a-1$ so the limit doesn't exist when $a$ is an integer. And it was correct to choose any value of $\delta$ like $1$ right? Can you help me when $a$ is not an integer looking at my opening post please?
It is not correct to choose any valiue of $\delta$, as the limit does not exists. For the not integer part. The function is constant outside the integer points. Do you know how to find $\lim_{x \to a} g(x)$ when $g(x)=n$ is the constant function?
@DanAsimov I wasn't giving a full formal explanation as I know how to do that and wanted to be quick. My question is how to compute the limit for when $a$ is not an integer. What have I done wrong / right ?
You need to give a fuller explanation of your reasoning, since it makes no sense at all to discuss the ε-δ definition of a limit without any mention of ε.
@DanAsimov For each $\epsilon > 0$, there must exist a $\delta > 0$ such that if $0 < |x-a| < \delta$, then $|f(x)-l|<\epsilon$. Happy? This is for when $a$ is not an integer as otherwise the definition is different for the one-sided limits.
@Dam there is an answer to this question and three dupes questions linked to it. The answer posted here is correct. As your notes. You can't use arbitrary $\delta$. e.g. $\delta=10$ does not work
@Marco So what I wrote in my opening post with $\delta = 1$ is incorrect? I set $n = \left\lfloor a\right\rfloor $ and get $n < a < n+1$. Whenever $0 < |x-a| < \delta $, $|f(x)-l|=|f(x)-n| = 0 < \epsilon$ where $\epsilon > 0$. Can you explain why as I don't understand?
Discussion on question by Dam: Floor …
Discussion on question by Dam: Floor …