Proving $\frac{x}{x-\lfloor \sin x \rfloor}$ has no limit as $x\to 0$ using the definition of limit I need to show that $$\lim_{x\to 0} \left(\frac{x}{x-\lfloor \sin x \rfloor} \right )$$
doesn't exist using the definition of limit (its negation).
I fail to choose $\varepsilon$ and $x$ correctly (obviously, $\varepsilon$ must depend on $L$; and $x$, perhaps, must depend on $\varepsilon$). Any suggestions?
 A: You can prove that $\lim_{x\to 0^+} \left(\frac{x}{x-\lfloor \sin x \rfloor} \right ) \neq \lim_{x\to {0^-}} \left(\frac{x}{x-\lfloor \sin x \rfloor} \right )$ as follows:
$$
\lim_{x\to {0^+}} \left(\frac{x}{x-\lfloor \sin x \rfloor} \right ) = \lim _{x \to 0^+}\frac{x}{x} = 1
$$
while,
$$
\lim_{x\to 0^-} \left(\frac{x}{x-\lfloor \sin x \rfloor} \right ) = \lim_{x \to 0^-}\frac{x}{x+1} = 0
$$
Edit: Having above explanation in mind, you can simply choose $\epsilon = \frac{1}{4}$, and either $x_1 = \min (\delta, \frac{1}{100})$ or $x_2 = -\min (\delta, \frac{1}{100})$ works because:
$$
f(x_1) = 1, \quad f(x_2) < 0
$$
and you can say $L$ cannot be chosen such that both $\left|f(x_1)-L\right|<\frac{1}{4}$ and $\left|f(x_2)-L\right|<\frac{1}{4}$ hold, because under that condition:
$$
\left| f(x_1)-f(x_2)\right| = \left| f(x_1)-L + L-f(x_2)\right|\leq \left|f(x_1)-L\right| + \left|f(x_2)-L\right|\leq \frac{1}{2}
$$
which is a contradiction.
A: Update
Let's face it: We all can see with our bare eyes that $\lim_{x\to0+}f(x)=1$ and $\lim_{x\to0-}f(x)=0$. Therefore the undertaking you propose is not an exercise in analysis, but an exercise in logic.
As your text shaded in beige says, we have to prove that for all $L\in{\Bbb R}$ something is true. Therefore let an $L\in{\Bbb R}$ be given. We have to produce an $\epsilon>0$ such that $\ldots\ $. In the case at hand the value $\epsilon:={1\over2}$ will do. I shall show that for all $\delta>0$ one can find a point $x$ with $|x-0|<\delta$ such that $|f(x)-L|\geq{1\over2}$. So let a $\delta>0$ be given., and choose an $n>1$ with $0<{1\over n}<\delta$.
We distinguish the cases (a) $L\geq{1\over2}$ and (b) $L<{1\over2}$. In case (a) put $x:=-{1\over n}$. Then
$$f(x)={-1/n\over-1/n-(-1)}=-{1\over n-1}<0\ ,$$
which implies $|f(x)-L|>{1\over2}$. In case (b) put $x:={1\over n}$. Then
$$f(x)={1/n\over1/n-0}=1\ ,$$
which implies $|f(x)-L|>{1\over2}$.
Original answer
Assume the limit $\lim_{x\to0}f(x)=:L$ exists.  Then there is a $\delta>0$ such that $|f(x)-L|<\epsilon:={1\over2}$ whenever $0<|x|<\delta$. Choose an $h$ with $0<h<\min\{\delta,1\}$. Then
$$|f(h)-f(-h)|\leq|f(h)-L|+|L-f(-h)|<1\ .$$
This contradicts the actual
$$f(h)-f(-h)={h\over h-0}-{-h\over -h-(-1)}={1\over1-h}>1\ .$$
