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?

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$$\lim_{x\to0^+}\frac{\sin(\lfloor x\rfloor+x)}{\lfloor x\rfloor+x}=1\ne\sin(1).$$
