Question

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

Answer 1

We know that the definition of modulus function. So, the modulus function of number x is defined as follows: \left| x \right|=\left\\{ \begin{aligned} & x,\text{ }x > 0 \\\ & -x,\text{ }x < 0 \\\ \end{aligned} \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{aligned} & x,\text{ }x > 0 \\\ & -x,\text{ }x < 0 \\\ \end{aligned} \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

& \Rightarrow \left| x \right|=1....(2) \\\ & \Rightarrow \left[ x \right]=1....(3) \\\ \end{aligned}$$ We know that $$\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{aligned} & \Rightarrow 1-x\text{ } < 1-1 \\\ & \Rightarrow 1-x\text{ } < 0 \\\ \end{aligned}$$ As the value of $$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{aligned} & \Rightarrow \left| 1-x \right|=-\left( 1-x \right) \\\ & \Rightarrow \left| 1-x \right|=x-1...(5) \\\ \end{aligned}$$ Now we should substitute equation (2), equation (3), equation (4) and equation (5) in equation (1). Then, we get $$\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{aligned} & \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{aligned}$$ From equation (7), it is clear that the value of L is equal to $$\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{aligned} & \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{aligned}$$ **From equation (8), it is clear that the value of L is equal to 0.** **Note:** Students may have a misconception that the modulus function of number x is defined as follows: $$\left| x \right|=\left\\{ \begin{aligned} & x,x < 0 \\\ & -x,x > 0 \\\ \end{aligned} \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.