I have the following limit to solve:
$$\lim_{x\to 0} \frac{\int_0^{x^2} (e^{t^2}-1)~dt}{\log(1+2x^6)}$$
The only thing I can do I guess is applying de l'Hôpital's rule and obtain:
- Numerator: $$ (e^{x^4} - 1)\times 2x $$
- Denominator: $$ \frac{12x^5}{1+2x^6} $$
What can I do now? Note that for the numerator I have used the theorem of integrals which says that the integral above (with $g(x)$ as upper bound and $h(dx)$ as lower bound) is:
$$ f(g(x))*g(x)' - f(h(x))*h(x)' $$
$\displaystyle \lim_{x\rightarrow 0}\frac{\int^{x^2}_{0}(e^{t^2}-1)dt}{\ln(1+2x^6)}$
applying de l'Hôpital's rule and $\displaystyle \lim_{y\rightarrow 0} \frac{e^y-1}{y} = 1$
$\displaystyle \lim_{x\rightarrow 0} \frac{(e^{x^4}-1)\cdot 2x}{\frac{12x^{5}}{1+2x^6}}=\lim_{x\rightarrow 0}2\frac{e^{x^4}-1}{x^4}\cdot \lim_{x\rightarrow 0} \frac{1+2x^6}{12} = 2\times \frac{1}{12}$