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.
