I'm wondering whether this limit can be evaluated.
I have no much experience in real analysis, so, looking for suggestions and intuition here.
So, I'm interested in computing the limit, or actually know what is the limit value of
$\lim_{x \to \infty , \epsilon\to 0} x^{1-\epsilon}$
My intuition is that may simply be indetermined
The double limit should be formalized as the following, assuming that the value is $L$:
For every $\eta>0$, there exist some $M>0$ and $\delta>0$ such that for every $x,\epsilon$ with $x\geq M$ and $0<|\epsilon|<\delta$, then $|x^{1-\epsilon}-L|<\eta$.
Take $\eta=1$, then we have for $x\geq M$ and $0<|\epsilon|<\delta$, then $|x^{1-\epsilon}-L|<1$. Pick an $\epsilon>0$ such that $\epsilon<\delta$ and that $1-\epsilon>0$, then $x^{1-\epsilon}<L+1$ for all such $x\geq M$. Taking $x\rightarrow\infty$ will make the latter inequality blowing up.