Let $f(x)=e^{\frac{1}{x}}$. In which of the following domains is it uniformly continuous?
a) (0,1)
b) (1,$\infty$)
c) (1,2)
My Attempt: For uniform continuity, we have a given $\epsilon$>0 and we need to find an $\delta$>0 such that whenever $|x-y|<\delta$, the following holds : $|f(x)-f(y)|<\epsilon$. In case (b) and (c), clearly the function $e^{\frac{1}{x}}$ is bounded. So we can always find such an $\delta$, where the above result holds true. But in case (a) when $x$ takes values very close to $0$, the function $f(x)$ diverges. So we can say it is not uniformly continuous.
Is my approach and justification correct? Can someone please give proper reasoning and mathematical explanation because I am not completely satisfied with what I have presented above.
Thanks in advance.
On intervals where your function has a Lipschitz constant (essentially, a bound on the first derivative) it will be uniformly continuous. Your function has a bounded derivative on both $(1,2)$ and $(1,\infty)$, so it is UC there. It is not UC on $(0,1)$ since you have arbitrarily large slopes close to $0$. In other words you can pick two points close to zero in such a way that $$ \frac{f(x_1)-f(x_2)}{x_1-x_2}\quad (*) $$ is as large as you want. Given an $\varepsilon$, no uniform $\delta$ will satisfy the definition of UC, you would have to choose a smaller and smaller $\delta$ the closer you get to zero, because of $(*)$.