Is $\lim_{t\to 0}\frac{e^{xt}-1}{x}=0$ uniformly on $x>0$?

Question

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?

Answer 1

If it is a uniform convergence it would converge to $0$ by pointwise convergence. So say $t< \delta \implies \sup_{x \in \mathbb{R}^+}|\frac{e^{tx}-1}{x}|<1$. That is clearly a contradiction.

Answer 2

For $x\ne 0$, our expression is equal to $t+x\frac{t^2}{2!}+x^2 \frac{t^3}{3!}+\cdots$. Note that for $|x|\lt 1$, this is smaller in absolute value than $|t|+\frac{|t|^2}{2!}+\frac{|t|^3}{3!}+\cdots$.

So there is absolutely no problem near $0$.

There is a problem for large $x$. For any fixed positive $t$, we can find $x$ such that $\frac{e^{xt}-1}{x}$ is arbitrarily large.

Thus we do not have uniform convergence in $(0,\infty)$. We do in $(0,M)$ for any positive $M$.