calculus - Why $lim_{xto0}frac{int_0^xfrac{t^2}{t^4+1}dt}{x^3}=lim_{xto0}frac{frac{x^2}{x^4+1}}{3x^2}$?

There is this limit question that I think we need to use the Fundamental Theorem of Calculus, but there is one little doubt during the process.

$$\lim_{x\to0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt$$

My doubt is are we allowed to differentiate on both numerator and denominator so that

$$\lim_{x\to0}\frac{\int_0^x\frac{t^2}{t^4+1}dt}{x^3}=\lim_{x\to0}\frac{\frac{x^2}{x^4+1}}{3x^2}$$

If we are allowed to do that, is it because

1. $\frac{\int_0^x\frac{t^2}{t^4+1}dt}{x^3}=\frac{\frac{x^2}{x^4+1}}{3x^2}$ (I mean we can differentiate both the numerator and denominator for all fractions in general) or




2. we can only do that when it involves limit as $x\to0$




3. In short, is there any theorem that says so (in general or just for some specific cases)?

Many thanks for the help!

Answer

$\lim_{x\to0}\dfrac{\int_0^x\frac{t^2}{t^4+1}dt}{x^3}$

$ = \lim_{x\to0}\dfrac{\frac{d}{dx}\int_0^x\frac{t^2}{t^4+1}dt}{\frac{d}{dx}x^3}$ by L'Hopital because $\int_0^x\frac{t^2}{t^4+1}dt \to 0$ as $x \to 0$

$ = \lim_{x\to0}\dfrac{\left(\frac{x^2}{x^4+1}\right)}{\frac{d}{dx}x^3}$ by the fundamental theorem of calculus

To be precise L'Hopital can only be used when the resulting limit exists, and it does as you can see later, but be sure to check in case it doesn't!