Question

For \mathop {\lim }\limits_{x \to 8} \dfrac{{\sin \left\\{ {x - 10} \right\\}}}{{\left\\{ {10 - x} \right\\}}} (where \left\\{ {} \right\\} denotes fractional part function),
(a) LHL exists but RHL does not exists
(b) RHL exists but LHL does not exists
(c) neither LHL or RHL exists
(d) both RHL and LHL exist and are equal to 1

Answer 1

Here, we need to find which of the given options is correct. We will use the definition of the fractional part function and find the left hand and right hand limits to check if they exist or not. A limit does not exist if substituting it makes the denominator of the expression equal to 0.

Complete step by step solution:
Rewriting \left\\{ {x - 10} \right\\} and \left\\{ {10 - x} \right\\}, we get
\left\\{ {x - 10} \right\\} = x - 10 - \left[ {x - 10} \right]
\left\\{ {10 - x} \right\\} = 10 - x - \left[ {10 - x} \right]
Thus, the function becomes
\Rightarrow \mathop {\lim }\limits_{x \to 8} \dfrac{{\sin \left\\{ {x - 10} \right\\}}}{{\left\\{ {10 - x} \right\\}}} = \mathop {\lim }\limits_{x \to 8} \dfrac{{\sin \left( {x - 10 - \left[ {x - 10} \right]} \right)}}{{\left( {10 - x - \left[ {10 - x} \right]} \right)}}
First, we will find the left hand limit of the given function.
The left hand limit of the function is
$LHL = \mathop {\lim }\limits_{x \to {8^ - }} \dfrac{{\sin \left( {x - 10 - \left[ {x - 10} \right]} \right)}}{{\left( {10 - x - \left[ {10 - x} \right]} \right)}}$
Since $x$ approaches 8 from below, the value of $x - 10$ lies between $- 2$ and $- 3$.
Therefore, the value of the function $\left[ {x - 10} \right]$ is $- 3$.
Similarly, since $x$ approaches 8 from below, the value of $10 - x$ lies between 2 and 3.
Therefore, the value of the function $\left[ {10 - x} \right]$ is 2.
Rewriting the function, we get
$LHL = \mathop {\lim }\limits_{x \to {8^ - }} \dfrac{{\sin \left( {x - 10 - \left( { - 3} \right)} \right)}}{{\left( {10 - x - 2} \right)}}$
Simplifying the expression, we get
$\begin{array}{l}LHL = \mathop {\lim }\limits_{x \to {8^ - }} \dfrac{{\sin \left( {x - 10 + 3} \right)}}{{8 - x}}\\\LHL = \mathop {\lim }\limits_{x \to {8^ - }} \dfrac{{\sin \left( {x - 7} \right)}}{{8 - x}}\end{array}$
If we substitute the limit $x = 8$ in the expression, then the denominator of the expression becomes 0.
Therefore, the left hand limit does not exist.
Next, we will find the right hand limit of the function.
The right hand limit of the function is
$RHL = \mathop {\lim }\limits_{x \to {8^ + }} \dfrac{{\sin \left( {x - 10 - \left[ {x - 10} \right]} \right)}}{{\left( {10 - x - \left[ {10 - x} \right]} \right)}}$
Since $x$ approaches 8 from above, the value of $x - 10$ lies between $- 1$ and $- 2$.
Therefore, the value of the function $\left[ {x - 10} \right]$ is $- 2$.
Similarly, since $x$ approaches 8 from above, the value of $10 - x$ lies between 1 and 2.
Therefore, the value of the function $\left[ {10 - x} \right]$ is 1.
Rewriting the function, we get
$RHL = \mathop {\lim }\limits_{x \to {8^ - }} \dfrac{{\sin \left( {x - 10 - \left( { - 2} \right)} \right)}}{{\left( {10 - x - 1} \right)}}$
Simplifying the expression, we get
$\begin{array}{l}RHL = \mathop {\lim }\limits_{x \to {8^ - }} \dfrac{{\sin \left( {x - 10 + 2} \right)}}{{9 - x}}\\\RHL = \mathop {\lim }\limits_{x \to {8^ - }} \dfrac{{\sin \left( {x - 8} \right)}}{{9 - x}}\end{array}$
If we substitute the limit $x = 8$ in the expression, then the denominator of the expression does not become 0.
Substituting the limit, we get
$\begin{array}{l}RHL = \dfrac{{\sin \left( {8 - 8} \right)}}{{9 - 8}}\\\RHL = \dfrac{{\sin 0}}{1}\\\RHL = 0\end{array}$
Therefore, the right hand limit does exist.

Thus, we get that the LHL does not exist, but the RHL exists. The correct option is option (b).

Note:
We used the fractional part function and the greatest integer to the left function in the solution. The fractional part function for a number $x$ can be written as \left\\{ x \right\\} = x - \left[ x \right], where $\left[ x \right]$ is the greatest integer to the left of $x$. For example: The fractional part of $2.1$ can be written as \left\\{ {2.1} \right\\} = 2.1 - \left[ {2.1} \right] = 2.1 - 2 = 0.1.