Question

Evaluate the limits. Applying L'Hopital's Rule Yield
lim┬(x→0)⁡〖1/x^3 〗 ∫_0^x▒〖t^2/(t^4+1) dt〗

Answer 1

$\lim\limits_{x\rarr0}$ $\frac{1}{x^3}$ $\int_{0}^x$ $\frac{t^2}{t^4+1}$$dt$ = $\lim\limits_{x\rarr0}$ $\frac{1}{3x^2}$$\frac{x^2}{x^4+1}$ = $\lim\limits_{x\rarr0}$ $\frac{1}{3(x^4+1)}$ = $\frac{1}{3}$

The limit of the initial fraction equals the limit of the fraction with derivative of numerator and derivative of denominator. Derivative of denominator is

$\frac{d}{dx}$ $(x^3)$ = $3x^2$ .

Derivative of numerator is

$\frac{d}{dx}$ $\int_{0}^x$ $\frac{t^2}{t^4+1}dt$ = $\frac{d}{dx}$ $(\Phi(x)-\Phi(0))$= $\frac{x^2}{x^4+1}$ .

Here $\Phi(x)$ is antiderivative of $\frac{t^2}{t^4+1}$ in point $x$.