The value of $\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \log (1+t)}{t^4+4} d t$ is

Question

Aicharya · Accepted Answer

$\lim _{x ightarrow 0} \frac{1}{x^3} \int_0^{x t} \frac{\log (1+t)}{4+t^4} d t$ Using L' Hospital's rule, $ \begin{aligned} & =\lim _{x ightarrow 0} \frac{\frac{x \log (1+x)}{4+x^4}}{3 x^2} \ = & \lim _{x ightarrow 0} \frac{\log (1+x)}{3 x} \cdot \frac{1}{4+x^4} \ = & \frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}\left[	ext { using, } \lim _{x ightarrow 0} \frac{\log (1+x)}{x}=1ight] \end{aligned} $

The value of $\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \log (1+t)}{t^4+4} d t$ is

Solution