Limits of functions that can't be attacked by Taylor series or L'hopital's rule

There is no hope for a limit in this problem. First, to get rid of the obvious obstacle, we consider the expression only for $x\in A = \mathbb R \setminus \pi\mathbb Z.$ Then we have a bona-fide function $f(x)$ on $A.$ At the endpoints of the intervals $(1/((n+1)\pi), 1/(n\pi))$ we have $f \to \infty.$ Thus $f$ is unbounded in $A\cap (0,\delta)$ for each $\delta > 0.$ This already shows $f$ can't have a finite limit. Could the limit be $\infty?$ No, because for each $n$ there is a unique $x_n> 0$ such that $x_n+e^{-1/x_n^2} = 1/(n\pi).$ By the intermediate value theorem, it follows that for every $\delta > 0,$ $f(A \cap (0, \delta)) =[0,\infty).$