Problem: Prove that $f(x)=e^{\frac{1}{x}}$ is uniformly continuous on $(a,1)$ for $a \in (0,1)$.
My solution: By using the Lagrange theorem for $x,y \in (a,1)$ we have $$\frac{f(x)-f(y)}{x-y}=f'(c)=e^{-\frac{1}{c^2}},$$ where $c \in (x,y)$. This yields
$$\vert e^{\frac{1}{x}}-e^{\frac{1}{y}} \vert < e^{-1} \vert x-y \vert.$$
So, $\forall\varepsilon > 0$, $\exists \delta =\varepsilon e >0$ that $\forall x,y \in (a,1)$, if $$\vert x-y \vert < \delta$$ then $$\vert f(x) -f(y) \vert = \vert e^{\frac{1}{x}}-e^{\frac{1}{y}} \vert < e^{-1} \vert x-y \vert <e^{-1}\delta=\varepsilon.$$ Therefore $f$ is uniformly continous on (a,1).
Question: In my solution, I have not used "$x,y>a$". That means, $f$ is uniformly continous on $(0,1)$ but we can easily prove that $f$ is not uniformly continous on $(0,1)$. Could you show me why my solution is wrong?
$$ (e^{1/x})'=-\frac{1}{x^2}e^{1/x}, $$ and hence there exists a $c\in(x,y)$, such that $$ e^{1/x}-e^{1/y}=(x-y)\cdot \frac{1}{c^2}e^{1/c}. $$ Thus $$ |e^{1/x}-e^{1/y}|=|x-y| \frac{e^{1/c}}{c^2}\le |x-y| \frac{e^{1/a}}{a^2}. $$