Question

If [.] denotes, GIF, then $\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \left( {\left[ {\dfrac{{2018{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}} \right] + \left[ {\dfrac{{2020{\text{x}}}}{{{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}}}} \right]} \right) =$ ?
${\text{A}}{\text{. 4038}} \\\ {\text{B}}{\text{. 4037}} \\\ {\text{C}}{\text{. 4036}} \\\ {\text{D}}{\text{. 4039}} \\\$

Answer 1

Hint: In order to find the GIF of the given function, we try to simplify the term function by applying limits to the inverse trigonometric functions and simplifying by using the respective formula of each term and then we determine the GIF, i.e. greatest integer function values of the terms in the square brackets.

Complete step-by-step answer:
Given Data,
$\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \left( {\left[ {\dfrac{{2018{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}} \right] + \left[ {\dfrac{{2020{\text{x}}}}{{{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}}}} \right]} \right)$
Now we rewrite the given equation by taking the limit inside the terms as follows:
$\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \left( {\left[ {\dfrac{{2018{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}} \right] + \left[ {\dfrac{{2020{\text{x}}}}{{{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}}}} \right]} \right)$
$\Rightarrow \left( {2018\left[ {\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \dfrac{{{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}} \right] + 2020\left[ {\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \dfrac{{\text{x}}}{{{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}}}} \right]} \right) \\\ \Rightarrow \left( {2018\left[ {\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \dfrac{{{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}} \right] + 2020\left[ {\dfrac{1}{{\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \dfrac{{{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}}}} \right]} \right) \\\$
Now we know the formula of $\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \dfrac{{{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}} = 1$ and also the formula of $\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \dfrac{{{{\tan }^{ - 1}}{\text{x}}}}{{\text{x}}} = 1$
Substituting these values in the above equation we get,
$\Rightarrow \left( {2018\left[ 1 \right] + 2020\left[ {\dfrac{1}{1}} \right]} \right)$
⟹2018 + 2020 = 4038.

Therefore, $\mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \left( {\left[ {\dfrac{{2018{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}} \right] + \left[ {\dfrac{{2020{\text{x}}}}{{{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}}}} \right]} \right) =$ 4038
Option A is the correct answer.

Note: In order to solve this type of question the key is to know the formulae of limits of the inverse trigonometric functions. The greatest integer function gives us the greatest possible value of the integer inside when applied to a limit, if there is a constant inside it the value is the same.
The same question can alternatively be solved in a different way using L –Hospital’s rule as follows:
The given function is in $\dfrac{0}{0}$ , that is when the limit x → 0 is applied both the numerator and the denominator of both terms becomes zero. So we can use L – hospital rule:
$\Rightarrow \mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \left( {\left[ {\dfrac{{2018{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}}}{{\text{x}}}} \right] + \left[ {\dfrac{{2020{\text{x}}}}{{{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}}}} \right]} \right) \\\ \Rightarrow \mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \left( {2018\left[ {\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{si}}{{\text{n}}^{ - 1}}{\text{x}}} \right)}}{{\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\text{x}} \right)}}} \right] + 2020\left[ {\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\text{x}} \right)}}{{\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right)}}} \right]} \right) \\\ \Rightarrow \mathop {{\text{Lt}}}\limits_{{\text{x}} \to {\text{0}}} \left( {2018\left[ {\dfrac{{\dfrac{1}{{\sqrt {{\text{1 - }}{{\text{x}}^2}} }}}}{1}} \right] + 2020\left[ {\dfrac{1}{{1 + {{\text{x}}^2}}}} \right]} \right) \\\ \Rightarrow \left( {2018\left[ {\dfrac{{\dfrac{1}{{\sqrt {{\text{1 - 0}}} }}}}{1}} \right] + 2020\left[ {\dfrac{1}{{1 + 0}}} \right]} \right) \\\$
⟹2018 + 2020 = 4038