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

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?

• Fix any $t > 0$. See whether $x \mapsto \frac{e^{tx}-1}{x}$ is bounded or not. – Daniel Fischer Jul 14 '13 at 16:52
• Uniform in what sense? – Pedro Tamaroff Jul 14 '13 at 16:52
• @PeterTamaroff I apologize for the ambiguity; I've clarified the question. – Eric Auld Jul 14 '13 at 16:56
• @DanielFischer Thanks, I think that settles that it is not uniform! – Eric Auld Jul 14 '13 at 17:02

## 2 Answers

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.

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 $M$.

• Great, so it is uniform on $(0,1]$. I think the problem may lie when $x$ is large, as Daniel Fischer suggests. – Eric Auld Jul 14 '13 at 17:00
• @EricAuld The convergence is uniform on every $(0,\, T]$, but not on all of $(0,\, \infty)$. Locally uniform convergence is sufficient for many purposes. – Daniel Fischer Jul 14 '13 at 17:02