The value of $\lim_{x\to 0} \frac{1}{x^{3}} \int_{0}^{x} \frac{t \ln(1+t)}{t^{4}+4}\,dt$ is

The value of lim_(x→ 0) 1/x³ ∫₀ˣ (t ln(1+t))/(t⁴+4) dt is $\lim_{x\to 0} \dfrac{1}{x^{3}} \int_{0}^{x} \dfrac{t\ln(1+t)}{t^{4}+4}\,dt$ $\text{To find: the limiting value}$ $\text{Plan: } 0/0 \text{ form} \Rightarrow \text{L'Hôpital}$ $L = \lim_{x\to 0}\dfrac{F'(x)}{3x^{2}},\quad F(x) = \int_{0}^{x} g(t)\,dt$ $F(x) = \int_{0}^{x} \dfrac{t\ln(1+t)}{t^{4}+4}\,dt,\qquad L = \lim_{x\to 0}\dfrac{F(x)}{x^{3}}$ $F(0) = 0,\quad x^{3}\to 0 \Rightarrow \text{apply L'Hôpital}$ $L = \lim_{x\to 0}\dfrac{F'(x)}{3x^{2}} = \lim_{x\to 0}\dfrac{g(x)}{3x^{2}}$ $g(x) = \dfrac{x\ln(1+x)}{x^{4}+4}$ $L = \lim_{x\to 0}\dfrac{\ln(1+x)}{3x\,(x^{4}+4)}$ $L = \lim_{x\to 0}\left[\dfrac{\ln(1+x)}{x}\cdot\dfrac{1}{3(x^{4}+4)}\right] = 1\cdot\dfrac{1}{3\cdot 4} = \dfrac{1}{12}$ $g(t) \approx \dfrac{t\cdot t}{4} = \dfrac{t^{2}}{4} \Rightarrow \int_{0}^{x} g(t)\,dt \approx \dfrac{x^{3}}{12}$ $\Rightarrow L \approx \dfrac{x^{3}/12}{x^{3}} = \dfrac{1}{12}$ $\boxed{\text{option (2)}}$