Please tell me how to calculate this limit:
$$
\lim_{x \to 0}\frac{2x^7-\displaystyle{\int_{x^2}^{x^3} \text{sin}(t^2)dt}}{\text{tg}(x^6)}
$$
I want to apply L'Hopital's rule, but the problem is that the numerator contains an integral with variable limits…
Try expanding the integral in a series. Or apply the mean value theorem for the integral? In this case, does the one-sided nature of the limit affect?
Best Answer
As an alternative to l'Hospital by series expansion
$$\int_{x^2}^{x^3} \text{sin}(t^2)dt=-\frac{x^6}3+o(x^6)$$
then
$$\frac{2x^7-\displaystyle{\int_{x^2}^{x^3} \text{sin}(t^2)dt}}{\text{tg}(x^6)}=\frac{2x^7+\frac{x^6}3+o(x^6)}{x^6+o(x^6)}=\frac{2x+\frac{1}3+o(1)}{1+o(1)}\to \frac 13$$