# 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.

• Is there any chance that it said $\dfrac{\sin \lfloor x+1\rfloor } {x+1}$? $\qquad$ – Michael Hardy Jul 6 '16 at 17:42
• No, the question is correct – Aakash Kumar Jul 6 '16 at 17:49
• Well, the value of f(x) when $-1\leq x<0$ is defined to be 0 – Ariana Jul 6 '16 at 17:53
• The function is $0$ immediately to the right of $-1$. – André Nicolas Jul 6 '16 at 17:53

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 $$\lim_{x\to0^-}g(x)=\sin1$$