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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 I can pick any value of $\delta$ (I choose $\delta=1$).

However, when $a$ is not an integer I am struggling. 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 $\epsilon > 0$.

Does this mean I can again choose whatever $\delta$ value I want (like $1$ for example)? My notes say that $\delta = \text{min}\{a-n,n+1-a\}$

Dam
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1 Answers1

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If $n = \left \lfloor{a}\right \rfloor $, then by letting $\delta = min\{a-n, n+1-a\}$, this guarantees that $x$ is on the same "floor" as $a$. Now, $|f(x) - n| = |\left \lfloor{x}\right \rfloor - n|$. Since $|x-a|<\delta$, $\left \lfloor{x}\right \rfloor = n $, so $|f(x) - n| =0 <\epsilon$. Thus, for any $\epsilon>0$, there exists a $\delta>0$ such that $|x-a|<\delta$ $\implies$ $|f(x)-n|<\epsilon$.

Leaf
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