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.
Best Answer
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.