Question

Find the value of the limit $\mathop {\lim }\limits_{x \to 1} \dfrac{{x\sin \left( {x - [x]} \right)}}{{x - 1}}$, where [.] denotes the greatest integer function.
A.0
B.-1
C.Non-existence
D.None of these

Answer 1

Hint: In this question, we need to evaluate the value of the given function such that [.] denotes the greatest integer function. For this, we will use the relation of the left-hand limit and the right-hand limit for the given function.

Complete step-by-step answer:
To determine the limit of the function $\mathop {\lim }\limits_{x \to 1} \dfrac{{x\sin \left( {x - [x]} \right)}}{{x - 1}}$, we need to evaluate the left-hand limit and the right-hand limit of the function and if the result is equal then, equate the result to determine the resultant value.
The value of the left-hand limit of the function at x=1 is defined as $f\left( {{x^ - }} \right)$ so, the value of the function at $x = 1 - h$ where h is infinitesimally small and is tending to zero is given by
$
  f\left( {{x^ - }} \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {\dfrac{{x\sin \left( {x - [x]} \right)}}{{x - 1}}} \right) \\
   = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{(1 - h)\sin \left( {(1 - h) - [1 - h]} \right)}}{{(1 - h) - 1}}} \right) \\
 $
As, the value of h is infinitesimally small so, the greatest integer of the term [1-h] will be zero. So,
$
  f\left( {{x^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{(1 - h)\sin \left( {(1 - h)} \right)}}{{ - h}}} \right) \\
   = \left( {\dfrac{{\sin \left( 1 \right)}}{{ - 0}}} \right) \\
   = \infty - - - - (i) \\
 $
The value of the right-hand limit of the function at x=1 is defined as $f\left( {{x^ + }} \right)$ so, the value of the function at $x = 1 + h$ where h is infinitesimally small and is tending to zero is given by
$
  f\left( {{x^ + }} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\dfrac{{x\sin \left( {x - [x]} \right)}}{{x - 1}}} \right) \\
   = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{(1 + h)\sin \left( {(1 + h) - [1 + h]} \right)}}{{(1 + h) - 1}}} \right) \\
 $
As, the value of h is infinitesimally small so, the greatest integer of the term [1+h] will be one. So,
$
  f\left( {{x^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{(1 + h)\sin \left( {(1 + h) - 1} \right)}}{h}} \right) \\
   = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\sin \left( h \right)}}{h}} \right) \\
   = 1 - - - - (ii) \\
 $
From the equations (i) and (ii) we can see that the value of the functions at the left-hand limit is equal to the value of the function at the right-hand limits so, the given function is not continuous at x=1.
Hence, the limits of the function $\mathop {\lim }\limits_{x \to 1} \dfrac{{x\sin \left( {x - [x]} \right)}}{{x - 1}}$ do not exist.

So, the correct answer is “Option C”.

Note: Students must note here that for the limit of the function to exists, the left-hand limit of the function must be equal to the right-hand limit of the function. Here, as the left-hand limit does not equal to the right-hand limit, then we can say that the limit does not exist at all.