Question

The value of the given integral $\mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{x^3}}}\int_0^x {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}dt} $ is
$\left( a \right)0$
$\left( b \right)\dfrac{1}{{12}}$
$\left( c \right)\dfrac{1}{{24}}$
$\left( d \right)\dfrac{1}{{64}}$

Answer 1

Hint: In this particular question use the concept that if the limit is in the form of $\dfrac{0}{0}$ then we use L’ hospitals’ rule i.e. differentiate the numerator and denominator separately so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given limit
$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{x^3}}}\int_0^x {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}dt} $
Let, $L = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{x^3}}}\int_0^x {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}dt} $
$ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\int_0^x {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}dt} }}{{{x^3}}}$
Now when we substitute x = 0 in the limit we get $\dfrac{{\int_0^0 {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}dt} }}{{{0^3}}} = \dfrac{0}{0}$ so it is called as indeterminate form so we use L’ hospitals’ rule i.e. differentiate the numerator as well as the denominator we get,
$ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\int_0^x {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}dt} }}{{\dfrac{d}{{dx}}{x^3}}}$
Now as we know that according to Leibniz integral rule, $\dfrac{d}{{dx}}\left( {\int\limits_a^b {g\left( x \right)dx} } \right) = {\left( {g\left( x \right)} \right)_{x = b}} - {\left( {g\left( x \right)} \right)_{x = a}}$ so according to this property differentiate the above equation and we also know that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ we have,
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}} \right)}_{t = x}} - {{\left( {\dfrac{{t\ln \left( {1 + t} \right)}}{{{t^4} + 4}}} \right)}_{t = 0}}}}{{3{x^2}}}\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{x\ln \left( {1 + x} \right)}}{{{x^4} + 4}}} \right) - \left( {\dfrac{{0\ln \left( {1 + 0} \right)}}{{{0^4} + 4}}} \right)}}{{3{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{x\ln \left( {1 + x} \right)}}{{{x^4} + 4}}} \right)}}{{3{x^2}}}\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{{3x\left( {{x^4} + 4} \right)}}\]
Now the above limit is also written as
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\ln \left( {1 + x} \right)}}{x}}}{{3\left( {{x^4} + 4} \right)}}\]
Now as we all know that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x} = 1$ so use this property we have,
\[ \Rightarrow L = \dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x}}}{{\mathop {\lim }\limits_{x \to 0} 3\left( {{x^4} + 4} \right)}} = \dfrac{1}{{3\left( {0 + 4} \right)}} = \dfrac{1}{{12}}\]
So this is the required value of the limit.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember the Leibniz integral rule, the differentiation of definite integral is given as $\dfrac{d}{{dx}}\left( {\int\limits_a^b {g\left( x \right)dx} } \right) = {\left( {g\left( x \right)} \right)_{x = b}} - {\left( {g\left( x \right)} \right)_{x = a}}$, and always recall the basic limit property that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x} = 1$, so first differentiate then use the basic limit property and then simplify we will get the required value of the limit.