Fix $\varepsilon>0$, we want to find $\delta>0$ such that $|x-x_0|<\delta$ implies $|h(x)-h(x_0)|<\varepsilon$. We have \[ \left|h(x)-h(x_0)\right|= \left|\int_0^x e^{t^2}\mathrm{d}t - \int_0^{x_0} e^{t^2}\mathrm{d}t\right|=\left|\int_{x_0}^x e^{t^2}\mathrm{d}t\right|. \] Since we know $e^{t^2}$ is continuous and $1$ is integrable, by the mean-value theorem for integrals we have a $c\in[a,b]$ such that \[ \left|\int_{x_0}^x e^{t^2}\mathrm{d}t\right| = e^{c^2}\left|\int_{x_0}^x\mathrm{d}t\right| = e^{c^2}|x-x_0| < \varepsilon \] if we put $\delta = \varepsilon/e^{c^2}$.