I would like to determine whether the following limit is uniform on $x\in (0,\infty)$: $$\lim_{t\to 0}\frac{e^{xt}-1}{x}.$$
By "uniform" here, we mean $\exists \delta_\epsilon>0$ such that $\frac{e^{xt}-1}{x}<\epsilon$ for all $0<t<\delta_\epsilon$, for all $x>0$.
From the Taylor remainder we can see that there is a point $t^*$ between $0$ and $t$ such that $\frac{e^{xt}-1}{x}=\frac{xt + x^2 (t^*)^2}{x}=t + x(t^*)^2$. This might suggest that the limit is not uniform, but of course we don't know the manner in which $t^*\to 0$. Investigating the derivative with respect to $t$, it is $e^{xt}$. Again we might try the Taylor estimate here, but that doesn't seem to be getting me anywhere.
Any ideas?