Question

For each \[t \in R\], let [t] be the greatest integer less than or equal to \[t\]. Then,
\[\mathop {\lim }\limits_{x \to 1 + } \dfrac{{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {\dfrac{\pi }{2}[1 - x]} \right)}}{{\left| {1 - x} \right|[1 - x]}}\]
A.Equals 1
B.Equals 0
C.Equals \[ - 1\]
D.Does not exist

Answer 1

Hint: Here we have to put the value of the \[x\] in the equation such that its limit will change. Then we will solve the equation accordingly using the limits. We will find the value of greatest integer and substitute it in the equation so that we will get the required answer.

Complete step by step solution:
Firstly, we will put the value of \[x\] in the equation such that its value of the limit of the equation will change and will tend to\[0\].
So from the equation we get that the limit of \[x\] tends to \[1 + \]. Therefore we can write it as
\[x = 1 + h\] and by we can say that value of \[h\] tends to\[0\]. Therefore by putting this in the given equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {1 - \left| {1 + h} \right| + \sin \left| {1 - \left( {1 + h} \right)} \right|} \right)\sin \left( {\dfrac{\pi }{2}[1 - \left( {1 + h} \right)]} \right)}}{{\left| {1 - \left( {1 + h} \right)} \right|[1 - \left( {1 + h} \right)]}}\]
Simplifying the above equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {1 - 1 - h + \sin \left| {1 - 1 - h} \right|} \right)\sin \left( {\dfrac{\pi }{2}[1 - 1 - h]} \right)}}{{\left| {1 - 1 - h} \right|[1 - 1 - h]}}\]
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( { - h + \sin \left| { - h} \right|} \right)\sin \left( {\dfrac{\pi }{2}[ - h]} \right)}}{{\left| { - h} \right|[ - h]}}\]………………. (1)
Now it is given in the question that \[\left[ t \right]\] be the greatest integer less than or equal to \[t\]. So we have to find the value of greatest integer of \[[ - h]\] so that we will put its value in equation (1) and solve it further. We know that if the value of \[x\] vary in the interval of \[n \le x < \left( {n + 1} \right)\] then its greatest integer value is \[[x] = n\].
So we know that the value of h will vary in the interval of\[0 \le h < 1\]so the value of the negative h will vary in the interval of\[ - 1 \le - h < 0\]. Therefore the greatest integer value of\[[ - h]\]will be\[ - 1\].
Now we will put the value of\[[ - h]\]in the equation (1) and solve further. So, we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( { - h + \sin \left| { - h} \right|} \right)\sin \left( {\dfrac{\pi }{2}( - 1)} \right)}}{{\left| { - h} \right|( - 1)}}\]
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( { - h + \sin h} \right)\sin \left( {\dfrac{\pi }{2}( - 1)} \right)}}{{h( - 1)}}\]
We know that the value of sin function at negative \[\dfrac{\pi }{2}\]angle is\[ - 1\] as it is in the 4th quadrant and sin function is negative in the 4th quadrant. So, we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( { - h + \sin h} \right)( - 1)}}{{h( - 1)}}\]
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\sin h - h} \right)}}{h}\]
Now by simplifying the above equation we will get the value of the equation. Therefore
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin h}}{h} - 1 = 1 - 1 = 0\]
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {\dfrac{\pi }{2}[1 - x]} \right)}}{{\left| {1 - x} \right|[1 - x]}} = 0\]
Hence, the value of the equation is equal to 0.
\[\therefore \] Option B is the correct option.

Note: Here we have to note that in the questions where the limit function is available we have to make the equation in such a way that the limit of the equation will become 0 or 1. This is because it will make the calculation very easy. We should also know that the mod function is the function which will return only the positive value of the function present inside the mod function.