Prove that a limit doesn't exist In my assignment I have to prove that the following function doesn't have a limit:
$$f(x)=\frac{x}{x-\lfloor{\sin x}\rfloor}$$ when $${x \to 0}$$
In other words, I have to prove that for every $\delta$ such that $0<|x-{x_0}|<\delta,$ $|f(x)-L|>\epsilon$.
In fact, I don't even know how to start. Do I choose my $\delta$? I know that the sin function has certain values that's supposed to help me. How do I use it?
I have to prove this by definition, so I can't use limit arithmetic for example
Thanks.
 A: The first thing I try to do in situations like this is to sketch a graph of the function in the neighborhood where the limit is evaluated.  In your case, it is not hard to see that when $x$ is sufficiently small and positive, $\lfloor \sin x \rfloor = 0$, for example.  So then $$f(x) = \frac{x}{x- \lfloor \sin x \rfloor}$$ in this region of $x$ will simply be equivalent to $x/x = 1$.
But for $x < 0$ but sufficiently close to $0$ (e.g., $x = -0.1$), $\lfloor \sin x \rfloor$ is no longer zero, but $-1$, because you are "rounding down."  Therefore, $f(x) = x/(x+1)$ in this region, and you can sketch the graph accordingly.
Now you can see that there is a jump discontinuity at $x = 0$.  Moreover, you can see that the discontinuity is of size $1$, which is formally represented by the fact that $$\left| \lim_{x \to 0^+} f(x) - \lim_{x \to 0^-} f(x) \right| = 1.$$  That suggests a choice of $\epsilon < 1$ will work to prove the two-sided limit does not exist, since if, say, $\epsilon = 1/2$, you will not be able to find any $\delta$ such that the difference $|f(x) - L| < 1/2$ in the neighborhood of $x = 0$.  To formalize this reasoning is an exercise I leave to you.
A: In the definition of "limit" choose $\delta = \frac14$.  Then we will show that there is no combination of limit value $v$ and "small enough" $\epsilon > 0$ such that whenever $|x-0| < \epsilon$, 
$$\left| \frac{x}{x - \lfloor \sin x \rfloor} - v \right| < \delta
$$
If we show this, then the definition of a limit cannot be met and $f(x)$ has no limit as $x\to 0$.
First, consider positive $x$, and choose a "test" $x$ which is less than $\epsilon$ and also less than $\pi/4$.  For such an $x$,  $\lfloor \sin x \rfloor = 0$ and $f(x) = 1$.  Note that within any $\epsilon$-neighborhood about zero, there are such values of $x$.  Thus if a suitable $v$ exists, we must have 
$$
\frac{3}{4} < v < \frac{5}{4}
$$
Next, consider negitive $x$, and choose a negative "test" $x$ which is greater than   $-\epsilon$ and also greater than $-\pi/4$.  For such an $x$,  $\lfloor \sin x \rfloor = -1$ and $f(x) = \frac{x}{x+1}$.  But since $x > -\pi/4$, the denominator is positive and $f(x) < 0$.
(Again note that within any $\epsilon$-neighborhood about zero, there are such values of $x$.)
Thus if a suitable $v$ exists, we must have $v-\frac{1}{4} < 0$ or 
$$
v < \frac{1}{4}
$$
Since there are no numbers that are both less than $\frac14$ and at least $\frac34$, such a limit value $v$ cannot exist, and $f(x)$ has no limit.
