Animated Solution for Mathematics - Limits, Continuity and Differentiability: For each $t\in R$, let [t] be the greatest integer less than or equal to t. Then, ${\lim }_{x\to 1^{+}}\frac{(1-|x|+\sin |1-x|)\sin (\frac{\pi }{2}[1-x])}{|1-x|[1-x]}$

Welcome, future IITian! Today, we are going to dismantle a limit problem that looks like a monster, but beneath its terrifying exterior lies a beautiful, elegant structure.

When you see a limit involving absolute values and the greatest integer function, your first instinct might be panic. But let's take a deep breath and look at this through the lens of a mathematician.

The problem asks us to evaluate:

The very first thing we must do is understand the condition $x\to 1^{+}$. This tiny plus sign is our most important clue. It tells us that $x$ is approaching $1$ from the right side, meaning $x>1$.

This single piece of information allows us to strip away the absolute value and greatest integer functions, which are the primary sources of confusion in this problem.

Let's tackle the components one by one. First, consider $|x|$. Since $x>1$, $x$ is positive, so $|x|=x$.

Next, consider $|1-x|$. Since $x>1$, the expression $1-x$ is negative. The modulus function forces a negative expression to become positive by multiplying it by $-1$. Thus:

Finally, consider the greatest integer function $[1-x]$. As $x$ approaches $1$ from the right, $1-x$ approaches $0$ from the left, sitting in the interval $(-1,0)$.

The greatest integer less than or equal to any number in this interval is $-1$. So, $[1-x]=-1$. We have successfully tamed the beasts!

Now, let's substitute these values back into our limit:

Look at how the expression has collapsed. The term $\sin (\frac{\pi }{2}(-1))$ is just $\sin (-\frac{\pi }{2})$, which evaluates to $-1$.

Our expression now becomes:

The $(-1)$ in the numerator and the $(-1)$ in the denominator cancel out perfectly, leaving us with:

We are almost there. Let's split this fraction into two parts:

The first term, $\frac{1-x}{x-1}$, is simply $-1$.

The second term, ${\lim }_{x\to 1^{+}}\frac{\sin (x-1)}{x-1}$, is a classic standard limit. If we let $\theta =x-1$, then as $x\to 1^{+}$, $\theta \to 0^{+}$.

This becomes ${\lim }_{\theta \to 0^{+}}\frac{\sin \theta }{\theta }$, which we know is $1$. Adding these together, $-1+1=0$.

The monster has been defeated, and the result is a clean, beautiful $0$. Remember, in JEE Advanced, the most complex problems are often just simple concepts disguised in layers. Keep peeling back those layers, and you will always find the answer.