Calculus

$\displaystyle\lim_{x \to 0} \dfrac{\displaystyle\int_{0}^{x} \dfrac{t^2}{t^4+1}\cdot dt}{x^3}$

$\text{The Limit is of } \dfrac{0}{0} \text{ form. We can use L'Hospital's Rule} \\ \text{According to Leibniz integral rule : } \\ \text{If I } = \displaystyle\int_{g(x)}^{h(x)}f(t)\cdot dt \\ \text{ then } \quad \dfrac{dI}{dx} = f(h(x))\cdot h^{'}(x)-f(g(x))\cdot g^{'}(x) \\ \text{We have , } \\ \displaystyle\lim_{x \to 0} \dfrac{\dfrac{x^2}{x^4+1}\cdot 1- 0 }{3x^2} \\ \displaystyle\lim_{x \to 0} \dfrac{1}{3(x^4+1)} = \boxed{\dfrac13}$

Differentiate using Newton Leibnitz theorem for 0/0 form (L Hospital's rule)

You will get limit as 1/3(x^4+1)

As x tends to zero we get limit as 1/3