Question

Let $f:R \to R$ be a differentiable function having $f\left( 2 \right) = 6,f'\left( 2 \right) = \left( {\dfrac{1}{{48}}} \right)$. Then $\mathop {\lim }\limits_{x \to 2} \dfrac{{\int\limits_6^{f\left( x \right)} {\left[ {4{t^3}dt} \right]} }}{{\left[ {x - 2} \right]}} = $
1. $18$
2. $12$
3. $36$
4. $24$

Answer 1

Hint: In this question, we have to find the value of $\mathop {\lim }\limits_{x \to 2} \dfrac{{\int\limits_6^{f\left( x \right)} {\left[ {4{t^3}dt} \right]} }}{{\left[ {x - 2} \right]}}$. When we’ll put the limit answer will be in indeterminate form. So, start the solution using L’hospital’s Rule. Now to solve the numerator part apply Leibniz’s integral rule and put $f\left( 2 \right) = 6,f'\left( 2 \right) = \left( {\dfrac{1}{{48}}} \right)$ these values after putting the limit in the required answer.

Formula Used:
L’hospital’s Rule –
$\mathop {\lim }\limits_{x \to c} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
Leibniz’s integral rule –
$\dfrac{d}{{dx}}\left[ {\int\limits_a^b {f\left( t \right)dt} } \right] = f\left( b \right)\dfrac{d}{{dx}}\left( b \right) - f\left( a \right)\dfrac{d}{{dx}}\left( a \right)$

Complete step by step Solution:
$\mathop {\lim }\limits_{x \to 2} \dfrac{{\int\limits_6^{f\left( x \right)} {\left[ {4{t^3}dt} \right]} }}{{\left[ {x - 2} \right]}}$
Applying L’hospital’s Rule,
 $ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\dfrac{d}{{dx}}\left\{ {\int\limits_6^{f\left( x \right)} {\left[ {4{t^3}dt} \right]} } \right\}}}{{\dfrac{d}{{dx}}\left[ {x - 2} \right]}}$
$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{4{{\left[ {f\left( x \right)} \right]}^3}\dfrac{d}{{dx}}f\left( x \right) + 4{{\left( 6 \right)}^3}\dfrac{d}{{dx}}\left( 6 \right)}}{{\dfrac{d}{{dx}}\left[ {x - 2} \right]}}$
$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{4{{\left[ {f\left( x \right)} \right]}^3}f'\left( x \right)}}{1}$
$ = 4{\left[ {f\left( 2 \right)} \right]^3}f'\left( 2 \right)$
$ = 4 \times {\left( 6 \right)^3} \times \dfrac{1}{{48}}$
$ = 18$

Hence, the correct option is (1).

Note:The key concept involved in solving this problem is a good knowledge of Indeterminate form. Students must know that an indeterminate form is a two-function expression whose limit cannot be determined solely by the limits of the individual functions. For example, the given problem was in $\dfrac{0}{0}$ form, and to solve we applied L’hospital’s Rule. To calculate numerators in an easy way Leibniz’s integral rule was applied.