Evaluate the following limit- \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1-\left| x \right|+\sin \left| 1-x \right| \right)\left( \sin \dfrac{\pi }{2}\left[ 1-x \right] \right)}{\left| 1-x \right|\left[ 1-x \right]}\]

Hint: We know that the definition of modulus function. So, the modulus function of number x is defined as follows: \[\left| x \right|=\left\{ \begin{align}
  & x,\text{ }x > 0 \\
 & -x,\text{ }x < 0 \\
\end{align} \right.\]. Now, we should also know the definition of greatest integer function. So, the greatest integer function of a number x if \[n\le \text{ }x < \text{ }n+1\] where n is an integer is defined as follows: \[\left[ x \right]=n\]. We should also know about the L-Hospital rule to solve this problem. If the limit of a function \[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}\] is in the form of \[\dfrac{0}{0}\], then \[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}=\displaystyle \lim_{x \to a}\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\]. By using these concepts, we can find the value of \[\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{\left( 1-\left| x \right|+\sin \left| 1-x \right| \right)\left( \sin \dfrac{\pi }{2}\left[ 1-x \right] \right)}{\left| 1-x \right|\left[ 1-x \right]}\].

Complete step by step answer:
Let us assume the value of \[\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{\left( 1-\left| x \right|+\sin \left| 1-x \right| \right)\left( \sin \dfrac{\pi }{2}\left[ 1-x \right] \right)}{\left| 1-x \right|\left[ 1-x \right]}\] is equal to L.
\[\Rightarrow L=\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{\left( 1-\left| x \right|+\sin \left| 1-x \right| \right)\left( \sin \dfrac{\pi }{2}\left[ 1-x \right] \right)}{\left| 1-x \right|\left[ 1-x \right]}.....(1)\]
Now, we should know the definition of modulus function. So, the modulus function of number x is defined as follows: \[\left| x \right|=\left\{ \begin{align}
  & x,\text{ }x > 0 \\
 & -x,\text{ }x < 0 \\
\end{align} \right.\].
Now, we should also know the definition of greatest integer function. So, the greatest integer function of a number x if \[n\le \text{ }x < \text{ }n+1\] where n is an integer is defined as follows: \[\left[ x \right]=n\].
Now we should apply the definition of modulus function and greatest integer function in equation (1).
From the question, it is clear that the value of x is greater than 1.
So, the modulus value of \[\left| x \right|\] is equal to 1 as 1 is greater than 0. As x is greater than 1, the value of \[\left[ x \right]\] is equal to 1.
So, we get
\[\begin{align}
  & \Rightarrow \left| x \right|=1....(2) \\
 & \Rightarrow \left[ x \right]=1....(3) \\
\end{align}\]
We know that
\[\Rightarrow x\text{ } > 1\]
Now let us multiply with -1 on both sides.
\[\Rightarrow -x\text{ } < -1\]
Now we will add 1 on both sides.
\[\begin{align}
  & \Rightarrow 1-x\text{ } < 1-1 \\
 & \Rightarrow 1-x\text{ } < 0 \\
\end{align}\]
As the value of \[1-x\] is less than zero, then the value of \[\left[ 1-x \right]\] is equal to -1.
\[\Rightarrow \left[ 1-x \right]=-1.....(4)\]
By using the definition of modulus function, we can say that
\[\begin{align}
  & \Rightarrow \left| 1-x \right|=-\left( 1-x \right) \\
 & \Rightarrow \left| 1-x \right|=x-1...(5) \\
\end{align}\]
Now we should substitute equation (2), equation (3), equation (4) and equation (5) in equation (1). Then, we get
\[\Rightarrow L=\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{\left( 1-x+\sin \left( x-1 \right) \right)\left( \sin \dfrac{-\pi }{2} \right)}{-\left( x-1 \right)\left( -1 \right)}....(6)\]
Now let us substitute the value of x is equal to 1 in equation (6), then we get
\[\begin{align}
  & \Rightarrow L=\dfrac{\left( 1-1+\sin 0 \right)\left( -1 \right)}{-\left( 1-1 \right)\left( 1-1 \right)} \\
 & \Rightarrow L=\dfrac{\left( 0 \right)\left( -1 \right)}{0} \\
 & \Rightarrow L=\dfrac{0}{0}....(7) \\
\end{align}\]
From equation (7), it is clear that the value of L is equal to \[\dfrac{0}{0}\].
Now we should know the definition of L-Hospital. If the limit of a function \[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}\] is in the form of \[\dfrac{0}{0}\], then \[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}=\displaystyle \lim_{x \to a}\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\].
Now we should apply L-Hospital in equation (6).
\[\Rightarrow L=\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{\dfrac{d}{dx}\left( 1-x+\sin \left( x-1 \right) \right)\left( \sin \dfrac{-\pi }{2} \right)}{-\dfrac{d}{dx}\left( x-1 \right)\left( -1 \right)}\]
We know that the value of \[\sin \dfrac{-\pi }{2}\] is equal to -1.
\[\begin{align}
  & \Rightarrow L=\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{\dfrac{d}{dx}\left( 1-x+\sin \left( x-1 \right) \right)\left( \sin \dfrac{-\pi }{2} \right)}{-\dfrac{d}{dx}\left( x-1 \right)\left( -1 \right)} \\
 & \Rightarrow L=\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{-\dfrac{d}{dx}\left( 1-x+\sin \left( x-1 \right) \right)}{-\dfrac{d}{dx}\left( x-1 \right)\left( -1 \right)} \\
 & \Rightarrow L=\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{\dfrac{d}{dx}\left( 1-x+\sin \left( x-1 \right) \right)}{-\dfrac{d}{dx}\left( x-1 \right)} \\
 & \Rightarrow L=\displaystyle \lim_{x \to {{1}^{+}}}\dfrac{-1+\cos \left( x-1 \right)}{-1} \\
 & \Rightarrow L=\dfrac{-1+\cos 0}{-1} \\
 & \Rightarrow L=\dfrac{-1+1}{-1} \\
 & \Rightarrow L=0.....(8) \\
\end{align}\]

From equation (8), it is clear that the value of L is equal to 0.

Note: Students may have a misconception that the modulus function of number x is defined as follows: \[\left| x \right|=\left\{ \begin{align}
  & x,x < 0 \\
 & -x,x > 0 \\
\end{align} \right.\]. Students may also have a misconception that the greatest integer function of a number x if \[n\le x < n+1\] where n is an integer is defined as follows: \[\left[ x \right]=n+1\]. If this misconception, then we cannot get the correct answer.