<s>[INST] Evaluate the limit of $\frac{\sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor }$ at $x=-1$ $$f(x):=\begin{cases}\frac{ \sin \lfloor x+1\rfloor }{  \lfloor x+1\rfloor  }   &   \lfloor x+1\rfloor \ne0 \\
          0    &   \lfloor x+1\rfloor=0 \ \end{cases}$$
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.
 [/INST] 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
$$
\lim_{x\to0^-}g(x)=\sin1
$$
</s>