I want to compute $$\lim_{x\to 0}\int_0^x e^{t^2}-1 dt$$ I have thought that $f(t)=e^{t^2}-1$ is a continous function because it is composition of continous functions and so I can apply the Weierstrass theorem to bound the function from above and below. But I don't know exactly how I can do this...the Weiestrass theorem in fact states that a maximum and a minimum for a function exist in a closed and bounded interval. How can I can apply this here? Actually I would say $$\lim_{x\to 0}\int_{0}^x min\quad dt\leq \lim_{x\to 0}\int_0^x e^{t^2}-1 dt\leq \lim_{x\to 0}\int_{0}^x max\quad dt $$ but I don't know how I can motivate this...in which interval can I apply Weierstrass?
Moreover I have also thought that I can simply state that since when $x\to 0$ the lower and upper limits will coincide then I obtain $0$, but I am not so convinced...can you help me?