calculating the limit $\lim \limits_{x \to 6.12} \lfloor x \rfloor$ 
calculate the limit $\lim \limits_{x \to 6.12} \lfloor x \rfloor$


This is the solution in the book:
let $\delta = 0.1$ for all $x \in N_\delta^*(6.12)$ we have $6.02<x<6.22$ and $6 < x <7$ therefore $\lfloor x \rfloor = 6$. so the function $f(x)= \lfloor x \rfloor$ intersects with the function $g(x)=6$ in the neighborhood $N_\delta^*(6.12)$ as the Proposition in the book states : let $f$ and $g$ be function . if $\lim \limits_{x \to x_0} f(x)$ exists and there is a neighborhood of $x_0$ such that $f(x)=g(x)$ then $\lim \limits_{x \to x_0} f(x) = \lim \limits_{x \to x_0} g(x)$
according to that $\lim \limits_{x \to 6.12} \lfloor x \rfloor = \lim \limits_{x \to 6.12} 6 $
I tried solving differently but I am not sure if what I did is correct:
by definitions the limit exists if for all $\epsilon >0$ there exists $\delta >0$ such that for all $x$ that satisfies $0<|x-x_0|< \delta$ it satisfies $|f(x)-L|< \epsilon$
let $\epsilon >0$
$|\lfloor x \rfloor -6| < |x+1-6| = |x-5|$
and also $|x-6.12|=|x-5-1.12| \leq |x-5|+1.12$
let $x \in N_{0.1}(6.12)$ then $-0.1<x-6.12<0.1$ $\iff$ $6.09<x<6.22$ so $\lfloor x \rfloor = 6$
since $6.09<x<6.22$ we can subtract $5$ and get $1.09 <x-5 < 1.22$ so $\lfloor x \rfloor =1$
so from $|x-6.12|=|x-5-1.12| \leq |x-5|+1.12$ we get $|x-5|+1.12 < 1.22 <1.22\delta$ and as mentioned above
$|\lfloor x \rfloor -6| < |x+1-6| = |x-5| <1.22 \delta$ so I defined $1.22 \delta =\epsilon \iff \delta=\frac{\epsilon}{1.22}$
let $\delta=min \{\frac{\epsilon}{1.22},0.1\}$ and from here
$|\lfloor x \rfloor -6| < |x+1-6| = |x-5| <1.22 \delta$ so I defined $1.22 \delta =\epsilon \iff \delta=\frac{\epsilon}{1.22}$ $\implies$ $1.22 \cdot \frac{\epsilon}{1.22}= \epsilon$
In my way I tried assuming that the limit is $6$ and then proving that it actually exists by definition
Sorry for the translation I hope that it is understandable as it is hard to translate mathematical terms from different languages to English And thank you for any tips and help!
 A: Your proof is much more complicated than it needs to be. Here is a much shorter proof.
Let $\epsilon > 0$ be given and let $x \in N_{0.1}(6.12)$. Then, since $\lfloor x\rfloor=6,$ we get that
$$|\lfloor x \rfloor - 6| = |6-6| = 0 < \epsilon.$$
We have thus proven directly by definition that $\lim_{x \rightarrow 6.12} \lfloor x \rfloor = 6$.

A few remarks about your attempted proof. Apart from most of the calculations being unnecessary there are also some computational mistakes. For instance the inequality
$$|\lfloor x \rfloor - 6| < |x-5|$$
is not valid for all $x$. For instance we have that
$$|\lfloor 5.8 \rfloor - 6| = |5-6| = 1 > 0.8 = |5.8 - 5|.$$
Another inequality that doesn't work is
$$|x-5| + 1.12 < 1.22< 1.22\delta,$$
where the first inequality seems to be assuming that $|x-5| < 0.1$, which is not possible if you also want $|x-6.12|<0.1$, and the second inequality assumes $\delta > 1$, which is also contradicting, when you end up with $\delta \leq 0.1$.
I get that you are trying to express $\delta$ as a functional expression $\delta(\epsilon)$ depending on $\epsilon$. But what can actually be seen, is that $\delta = 0.1$ works for all $\epsilon > 0$, so you can actually choose $\delta(\epsilon)$ to be the constant function $0.1$.
