We are to evaluate the limit: $\underset{x\to 1+}{\mathrm{Lim}}\frac{(1-|x|+\sin |1-x\left|\right)\sin \left(\frac{\pi }{2}[1-x]\right)}{|1-x|[1-x]}$

Since $x\to 1+$, we have $x>1$. Let us define a new variable $h=x-1$. Then as $x\to 1+$, we have $h\to 0+$.

Now, substitute $x=1+h$ in the expression:

• $\left|x\right|=|1+h|=1+h$ (since $h>0$)
• $|1-x|=|1-(1+h\left)\right|=|-h|=h$
• $[1-x]=[1-(1+h\left)\right]=[-h]$

Now, since $h\to 0+$, we have $h>0$ but small. For example, if $0<h<1$, then $-h\in (-1,0)$. The greatest integer less than or equal to a number in $(-1,0)$ is $-1$. Therefore, $[-h]=-1$ for $0<h<1$.

So, we have:

• $[1-x]=[-h]=-1$
• $\sin \left(\frac{\pi }{2}[1-x]\right)=\sin \left(\frac{\pi }{2}\cdot (-1)\right)=\sin \left(-\frac{\pi }{2}\right)=-1$
• $1-\left|x\right|=1-(1+h)=-h$
• $\sin |1-x|=\sin \left(h\right)$
• So numerator becomes: $(-h+\sin (h\left)\right)\cdot (-1)=h-\sin \left(h\right)$
• Denominator becomes: $|1-x|\cdot [1-x]=h\cdot (-1)=-h$

Therefore, the expression becomes:

$\frac{h-\sin \left(h\right)}{-h}=-\frac{h-\sin \left(h\right)}{h}=-(1-\frac{\sin \left(h\right)}{h})$

Now, take the limit as $h\to 0+$:

$\underset{h\to 0+}{\mathrm{Lim}}-(1-\frac{\sin \left(h\right)}{h})=-(1-1)=-\left(0\right)=0$

Therefore, the limit is equal to 0.

Final Answer: equal 0

Related Topics

Limit Evaluation: The process of finding the value a function approaches as the input approaches some value. Techniques include substitution, simplification, and using standard limits like $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$.

Greatest Integer Function: Also known as the floor function, [x] denotes the greatest integer less than or equal to x. It is discontinuous at all integer points.

Absolute Value Function: |x| gives the non-negative magnitude of x. It is continuous everywhere but not differentiable at 0.

Formulae

Standard Limit: $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$

Behavior of [x]: For small positive h, [ -h ] = -1.