# Proving that the function $f(x)=e^\frac{-1}{x}$ is uniformly continuous in $(0,\infty)$

I got this problem:

Prove that the function $f(x)=e^\frac{-1}{x}$ is uniformly continuous in $(0,\infty)$ by the definition of uniform continuity.

I managed to prove it for $1\leq\epsilon$ as follows:
First we'll show that $\forall x_1,x_2\in (0,\infty), |e^\frac{-1}{x_2}-e^\frac{-1}{x_1}|<1$:

Let $x_1,x_2\in (0,\infty)$, Then because $0<x_1,x_2$, we get that $0<\frac{1}{x_1}, \frac{1}{x_2}$, which implies that $-\frac{1}{x_1}, -\frac{1}{x_2}<0$. Now because the function $e^x$ is increasing for all $x\in\mathbb{R}$, We get that $e^{-\frac{1}{x_1}}, e^{-\frac{1}{x_2}}<e^0=1$, And because $\forall x\in\mathbb{R}, 0<e^x$, We get that $0<e^{-\frac{1}{x_1}}, e^{-\frac{1}{x_2}}$, And so $0<e^{-\frac{1}{x_1}}, e^{-\frac{1}{x_2}}<1$.

Therefore, We get that $-1< -e^{-\frac{1}{x_1}}<0$, Which implies that $-1<e^{-\frac{1}{x_2}} - e^{-\frac{1}{x_1}}<1$, And so $|e^\frac{-1}{x_2}-e^\frac{-1}{x_1}|<1$, As was to be shown.

Now we'll prove that $\forall 1\leq \epsilon,\exists 0< \delta$ such that $\forall x_1,x_2\in(0,\infty)$, if $|x_2-x_1|<\delta$, then $|e^\frac{-1}{x_2}-e^\frac{-1}{x_1}|<\epsilon$:

Let $1\leq\epsilon$, And let's take $0<\delta=1$. Now we'll show that $\forall x_1,x_2\in(0,\infty)$, if $|x_2-x_1|<\delta$, then $|e^\frac{-1}{x_2}-e^\frac{-1}{x_1}|<\epsilon$:

Let $x_1,x_2\in(0,\infty)$ that satisfy $|x_2-x_1|<\delta$, then from what was shown before the proof we get that $|e^\frac{-1}{x_2}-e^\frac{-1}{x_1}|<1$, and because $1\leq \epsilon$, we get that $|e^\frac{-1}{x_2}-e^\frac{-1}{x_1}|<\epsilon$, As was to be shown.

Now I tried to prove that $\forall 0< \epsilon<1,\exists 0< \delta$ such that $\forall x_1,x_2\in(0,\infty)$, if $|x_2-x_1|<\delta$, then $|e^\frac{-1}{x_2}-e^\frac{-1}{x_1}|<\epsilon$.

But I didn't managed to proceed.

Thanks for any help.

It is a Lipschitz function, since the derivative is bounded (the derivative is continuous and is 0 as $x\to 0$ and $x \to \infty$), then you have a $L> 0$ s.t.
$$|x_1 - x_2 | \geq L | e^{\frac{-1}{x_1}} - e^{\frac{-1}{x_2}} | \, \, \, \forall x_1, x_2$$
So, for every $\epsilon > 0$, set $\delta =\epsilon L$ and if $|x_1 - x_2 | < \delta$
$$\epsilon = \delta / L > |x_1 - x_2 | / L \geq | e^{\frac{-1}{x_1}} - e^{\frac{-1}{x_2}} |$$