Let \[f:R \to R\] be a differentiable function having \[f\left( 2 \right) = 6\],
\[f'\left( 2 \right) = \left( {{1 \over {48}}} \right)\]. Then \[\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} \] equals :
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Official solution
\[\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt\]
\[ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}\]
This limit resembles a derivative because the fraction has the form 0/0 as x→2 since both the numerator (integral from 6 to f(x)) and the denominator (x−2) are zero when x=2.
Applying L′ Hospital rule
\[\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}\]
\[ = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)\]
\[ = 4 \times {6^3} \times {1 \over {48}} = 18\]