Animated Solution for Mathematics - Limits, Continuity and Differentiability: For each $x\in R$, let $[x]$ be the greatest integer less than or equal to $x$. Then ${\lim }_{x\to 0^{+}}\frac{x([x]+|x|)\sin [x]}{|x|}$ is equal to

The problem asks us to evaluate the limit:

The notation $x\to 0^{-}$ defines our universe as the interval $(-1,0)$. In this specific neighborhood, the behavior of the piecewise functions becomes rigid and predictable.

First, consider the Greatest Integer Function, $[x]$. By definition, it returns the largest integer less than or equal to $x$. For any $x$ in the interval $(-1,0)$, the integer immediately to the left is $-1$. Thus, we have:

Next, consider the Modulus Function, $|x|$. We know that $|x|=-x$ for all $x<0$. Since our interval is strictly negative, we substitute:

Now, we substitute these defined values back into our original expression:

We observe a common factor of $x$ in both the numerator and the denominator. Because the limit process implies $x$ approaches $0$ but never equals $0$, we are mathematically justified in canceling $x$:

Recall the odd function property of trigonometry, where $\sin (-\theta )=-\sin (\theta )$. Therefore, $\sin (-1)=-\sin (1)$. Substituting this into our expression yields:

Dividing by $-1$ effectively flips the signs of the numerator, transforming $(-1-x)$ into $(1+x)$. The expression simplifies to:

Since the expression is now continuous at $x=0$, we perform the final substitution:

The final result of the limit is $-\sin 1$.