$$f(x):=\begin{cases}\frac{ \sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor } & \lfloor x+1\rfloor \ne0 \\ 0 & \lfloor x+1\rfloor=0 \ \end{cases}$$ Then at $x=-1$ find the limit
My work
$$\lfloor x+1\rfloor=0$$ $$0\le\ x+1\lt1$$ $$f(x):=\begin{cases}\frac{ \sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor } & \lfloor x+1\rfloor \ne0 \\ 0 & -1\le x \lt 0 \ \end{cases}$$ $$LHL=\lim_{x\to -1^-}\frac{\sin(-1)}{-1}$$ $$RHL=\lim_{x\to -1^+}\frac{\sin(0)}{0}$$
$$LHL=\sin1$$ $RHL= \text{Not defined}$ But the answer say$$ RHL=0$$ Please tell me why I am wrong.
Read carefully the definition: the function is defined at the right of $-1$.
Make your life simpler and change $x+1$ into $x$, so you need the limits at $0$ of $$ g(x)=\begin{cases} \dfrac{\sin\lfloor x\rfloor}{\lfloor x\rfloor} & \text{if $\lfloor x\rfloor\ne0$} \\[6px] 0 & \text{if $\lfloor x\rfloor=0$} \end{cases} $$
For $0<x<1$, we have $\lfloor x\rfloor=0$, so $g(x)=0$ and therefore $$ \lim_{x\to0^+}g(x)=0 $$ For $-1<x<0$, we have $\lfloor x\rfloor=-1$ and so $g(x)=\frac{\sin(-1)}{-1}=\sin 1$; therefore $$ \lim_{x\to0^-}g(x)=\sin1 $$