Limit of an integral with an x border

How to evaluate this limit?

$$\lim_{x\to0}\frac{\int_{0}^{x}\sin(u^3)\;\text{du}}{x^4}$$

where the whole integral is under $x^4$

• can you use $$\LaTeX$$ please? Mar 23 '17 at 8:03
• is it $$\lim_{x->0}(\int_{0}^{x}sinu^3du)/x^4$$? Mar 23 '17 at 8:04
• the searched Limit should be $$\frac{1}{4}$$ Mar 23 '17 at 8:15
• Ok, i thought i used it. thanks for your help! Mar 23 '17 at 9:50

By the Fundamental Theorem of Calculus, write

$$\int_0^x \sin u^3\,du=F(x)-F(0)\;,\;\;\text{with}\;\;F'(t)=\sin t^3\implies$$

$$\lim_{x\to0}\frac{F(x)-F(0)}{x^4}\stackrel{\text{l'Hospital}}=\lim_{x\to0}\frac{\sin x^3}{4x^3}=\frac14$$

Note that both the numerator and the denominator approach zero as $x\to 0$. Thus, we can use L'Hôpital's Rule and take the derivative of both individually. The Fundamental Theorem of Calculus allows us to simplify the integral in the numerator.

$$\lim_{x\to 0}\frac{\int_0^x\sin u^3du}{x^4}=\lim_{x\to 0}\frac{\frac d{dx}\left(\int_0^x\sin u^3du\right)}{\frac d{dx}(x^4)}=\lim_{x\to 0}\frac{\sin x^3du}{4x^3}$$

Again, both numertor and denominator approach $0$. We can thus apply L'Hôpital's rule one more time.

$$\lim_{x\to 0}\frac{\sin x^3du}{4x^3}=\lim_{x\to 0}\frac{\frac d{dx}\left(\sin u^3du\right)}{\frac d{dx}(4x^3)}=\lim_{x\to 0}\frac{3x^2\cos x^3du}{12x^2}=\lim_{x\to 0}\frac 14\cos x^3=\frac 14$$

• very well explained. Thank you! Mar 23 '17 at 9:51