Will the limit exist? If $$f(x) = \begin{cases} \frac{\sin([x]+x)}{[x]+x} &\textrm{if } x \neq 0, \\ 1 &\textrm{if } x = 0, \end{cases}$$
where [.] is the greatest integer function, then does
$\lim_{x \to 0}f(x)$ exist or not?

I thought of using the property that  $\lim_{x \to 0} \frac{\sin g(x)}{g(x)}=1$ by which I thought that the limit should exist but my book said otherwise. What did I miss? Is there anything else that should be applied?
 A: The left and right limits at $0$ must be equal to $1$ for continuity. But you can see that as $x\to 0^-$, $$\frac{\sin([x] +x)}{[x] + x}= \frac{\sin(-1+x)}{-1+x} \to \sin 1 \ne 1 $$
A: What you missed was the fact that the function $x\mapsto\lfloor x\rfloor+x$ is discontinuous.
Not, that, if $x\in(-1,0)$, $\lfloor x\rfloor+x=x-1$ and that therefore$$\lim_{x\to0^-}\frac{\sin(\lfloor x\rfloor+x)}{\lfloor x\rfloor+x}=\frac{\sin(-1)}{-1}=\sin(1).$$But$$\lim_{x\to0^+}\frac{\sin(\lfloor x\rfloor+x)}{\lfloor x\rfloor+x}=1\ne\sin(1).$$
A: $\lim_{x \to 0} \frac{\sin g(x)}{g(x)}=1$ is only true if $g(x)$ is continuous and if $\lim_{x\to 0} g(x) = 0$.  Neither of those are true for $g(x) = [x]+x$.
.....
On the intervals $[0,1)$ and $(-1,0)$ then $[x]$ is a constant value.  If $0\le x < 1$ then $[x]= 0$.  And if $-1 < x < 0$ then $[x] = -1$.
$\lim_{x\to 0^+} f(x)$ if it exists will be equal to
$\lim_{x\to 0^+}\frac{\sin([x]+x)}{[x]+x} = \lim_{x\to 0^+}\frac {\sin (0 + x)}{0 + x} =\lim_{x\to 0}\frac {\sin x}x$ which be l'hopital is equal to $\lim_{x\to 0} \frac {\cos x}1 = 1$.
Meanwhile $\lim_{x\to 0^-} f(x)$ if it exists well be equal to
$\lim_{x\to 0^-}\frac{\sin([x]+x)}{[x]+x} = \lim_{x\to 0^-}\frac {\sin (-1 + x)}{-1 + x} =\frac {\sin (-1)}{-1}\ne 1$.
