
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 to
A. \[0\]
B. \[\sin 1\]
C. \[ - \sin 1\]
D. \[1\]
Answer
408k+ views
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.
Complete step by step answer:
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 \[\mathop {\lim }\limits_{x \to a} k\,f\left( a \right) = k\mathop {\lim }\limits_{x \to a} f\left( a \right)\].
Complete step by step answer:
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 \[\mathop {\lim }\limits_{x \to a} k\,f\left( a \right) = k\mathop {\lim }\limits_{x \to a} f\left( a \right)\].
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE
