# In which of the following domains is $e^{\frac{1}{x}}$ uniformly continuous?

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.

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 $$(*)$$.
• About your approach: being bounded is not enough to be able to choose a universal $\delta$. For example, take $f(x)=\sin(1/x)$ on $(0,1)$. This function is bounded but it is not uniformly continuous because it has oscillations with arbitrarily large frequency close to zero. What you need is to have a controlled slope. Your argument for part $a)$ seems correct: if your function has limit $\pm\infty$ at the end-point, it cannot be UC. – GReyes Oct 7 '19 at 3:14