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# For each $x \in R$, let $\left[ x \right]$ be the greatest integer less than or equal to $x$. Then, $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]}}{{\left| x \right|}}$ is equal toA. $0$B. $\sin 1$C. $- \sin 1$D. $1$

Last updated date: 19th Jul 2024
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Hint: Here in this question, we have to determine the given limit of a function having the greatest integer $[x]$ which is less than or equal to $x$. here the limit is left hand limit i.e., in the form of $\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)$ , in given function which ‘x’ approaches ‘0’ through values less than ‘0’. Then we apply the limit to the function and simplify by using the properties of limit function, we get the required solution.

A left-hand limit means the limit of a function as it approaches from the left-hand side.
The left hand limit of function $f\left( x \right)$ as $'x'$ tends to $'a'$ exists and is equal to ${l_2}$, if as $'x'$ approaches $'a'$ through values less than $'a'$. i.e.,
$\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = {l_2}$
Where, ${a^ - }$ means $\left( {a - h} \right)$ and $h \to 0$. Therefore, $f\left( {a - h} \right)$.
Consider the given limit function,
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]}}{{\left| x \right|}}$------(1)
Take some substitution for $\mathop {\lim }\limits_{x \to {a^ - }} f(x)$ put $x = a - h$ and change the limit as $x \to {a^ - }$ by $h \to 0$, then Equation (1) becomes
$\Rightarrow \,\,\,\mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {a - h} \right)\left( {\left[ {\left( {a - h} \right)} \right] + \left| {\left( {a - h} \right)} \right|} \right)\sin \left[ {\left( {a - h} \right)} \right]}}{{\left| {\left( {a - h} \right)} \right|}}$
But $a = 0$, then
$\Rightarrow \,\,\,\mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {0 - h} \right)\left( {\left[ {\left( {0 - h} \right)} \right] + \left| {\left( {0 - h} \right)} \right|} \right)\sin \left[ {\left( {0 - h} \right)} \right]}}{{\left| {\left( {0 - h} \right)} \right|}}$
$\Rightarrow \,\,\,\mathop {\lim }\limits_{h \to 0} \dfrac{{ - h\left( {\left[ { - h} \right] + \left| {\left( { - h} \right)} \right|} \right)\sin \left[ { - h} \right]}}{{\left| { - h} \right|}}$
As we know the greatest integer of $\left[ { - h} \right]$ is -1 i.e., $\left[ { - h} \right] = - 1$ and the absolute value of $\left| { - h} \right| = h$, then on substituting we have
$\Rightarrow \,\,\,\mathop {\lim }\limits_{h \to 0} \dfrac{{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)}}{h}$
On cancelling the like terms $h$ on both numerator and denominator, we have
$\Rightarrow \,\,\,\mathop {\lim }\limits_{h \to 0} \dfrac{{ - \left( { - 1 + h} \right)\sin \left( { - 1} \right)}}{1}$
By the product and quotient property of limit of a function, then
$\Rightarrow \,\,\,\dfrac{{\mathop {\lim }\limits_{h \to 0} \left( {1 - h} \right) \cdot \mathop {\lim }\limits_{h \to 0} \left( { - \sin 1} \right)}}{{\mathop {\lim }\limits_{h \to 0} \left( 1 \right)}}$
On applying a limit value, we get
$\Rightarrow \,\,\,\dfrac{{\left( {1 - 0} \right) \cdot \left( { - \sin 1} \right)}}{1}$
$\Rightarrow \,\,\, - \sin 1$
Hence, the left hand limit value of $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]}}{{\left| x \right|}} = - \sin 1$.

Therefore, option C is correct.

Note: Remember the function $\left[ x \right]$ is a greatest integer function for any real function. The function rounds -off the real number down to the integer less than the number. And the product and quotient properties of limits are defined as: The function $f\left( x \right)$ and $g\left( x \right)$ is are non-zero finite values, given that
$\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) \cdot g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)$
$\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}$ and
Also $\mathop {\lim }\limits_{x \to a} k\,f\left( a \right) = k\mathop {\lim }\limits_{x \to a} f\left( a \right)$.