JEE Advance - Mathematics (2010 - Paper 1 Offline - No. 14)
The value of $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int\limits_0^x {{{t\ln \left( {1 + t} \right)} \over {{t^4} + 4}}} dt$$ is
$$0$$
$${1 \over 12}$$
$${1 \over 24}$$
$${1 \over 64}$$
Explanation
$$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int_0^x {{{t\log (1 + t)} \over {4 + {t^4}}}dt} $$
Using L' Hospital's rule,
$$ = \mathop {\lim }\limits_{x \to 0} {{{{x\log (1 + x)} \over {4 + {x^4}}}} \over {3{x^2}}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over {3x}}\,.\,{1 \over {4 + {x^4}}}$$
$$ = {1 \over 3}\,.\,{1 \over 4} = {1 \over {12}}$$ [using, $$\mathop {\lim }\limits_{x \to 0} {{\log (1 + x)} \over x} = 1$$]
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