The value of $$\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^3}}}\int\limits_0^x {\frac{{t\,ln\left( {1 + t} \right)}}{{{t^4} + 4}}dt} $$     is-

Solution :
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^3}}}\int_0^x {\frac{{t\,ln\left( {1 + t} \right)}}{{{t^4} + 4}}dt\,\,\,\,\,\left[ {\frac{0}{0}{\text{form}}} \right]} $$
Applying L’ Hospital’s rule, we get
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{xln\left( {1 + x} \right)}}{{{x^4} + 4}}}}{{3{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{ln\left( {1 + x} \right)}}{x}.\frac{1}{{3\left( {{x^4} + 4} \right)}} \cr & = 1.\frac{1}{{12}} \cr & = \frac{1}{{12}} \cr} $$