Question

For each $x\in R$, let $\left[x\right]$ be the greatest integer less than or equal to $x$. Then $\underset{x\to 0{}^{-}}{lim}\frac{x\left(\right[x]+|x|)\sin \left[x\right]}{|x|}$ is equal to :

• A. $\sin 1$
• B. $0$
• C. $-\sin 1$
• D. $1$

Answer 1

The correct option is C $-\sin 1$

$\underset{x\to 0{}^{-}}{lim}\frac{x\left(\right[x]+|x|)\sin \left[x\right]}{|x|}$
We know,
As $x\to 0^{-},\left[x\right]=-1$ and $|x|=-x$

So, $\underset{x\to 0^{-}}{lim}\frac{x(-1-x)\sin (-1)}{-x}$

$=\underset{x\to 0^{-}}{lim}-\frac{(1+x)\sin 1}{1}$

$=-\frac{(1+0)\sin 1}{1}$

$=-\sin 1$