Analytic function with $f(1-1/n)=-n$ and $f(1+1/n)=-n$

Question

I have to answer the following question.

Can we find a function $f$ define on an open set $U$ which contains $1$ such that $f(1-1/n)=-n$ and $f(1+1/n)=n$

I know I have to use the identity theorem to conclude but it doesn't work here.

In fact, $z\mapsto \frac{-1}{1-z}$ would do the job, but is not define in $1$. Then i think such a function doesn't exist, any hint or idea to prove it ?

Answer 1

Since the limit $\lim_n-n$ does not exist (in $\mathbb C$) and since $\lim_n1-\frac1n=1$, there's no continuous function defined near $1$ satisfying that condition.