False warning messages in R, is it possible?

Question

I'm modeling GARCH-filtered standardized residuals via semiparametric distribution with Gaussian kernel and GPD (generalized pareto distribution) tails with thresholds at 5% and 95%. For some series I'm considering I have negative shape parameters $\xi$, which imply a finite left/right endpoint equal to $-\frac{\beta}{\xi}$.

Then, in the testing period (out-of-sample analysis) I use the quantile function to convert copula realizations into standardized residuals through the inverse CDF. I get the following warning:

Warning messages:

In log(1 + (xi * (as.vector(exceedances) - u))/beta) : Si è prodotto un NaN

NaNs are produced in the logarithm expression ($\xi$ shape parameter, $\beta$ scale parameter, $u$ threshold, $x$ value over the threshold). To check what that means, I solve $$1 + \xi \frac{(x-u)}{\beta}\leq0$$ when $\xi$ is negative $$x-u\geq-\frac{\beta}{\xi}$$ implying that somewhere in the future I have an excess over the threshold which exceeds the bound implied by negative $\xi$.

Since I manually verified that the message wasn't true (in the out-of-sample dataset I NEVER have standardized residuals which exceed the 5% and 95% thresholds by more than $-\frac{\beta}{\xi}$), I decided to verify this with R using this code:

p <- 0
for (i in 1:2){
  for (j in 1:9){
    for (z in 1:100){
      if (is.nan(ronaldo[i,j,z])==TRUE) {p=1}
    }
  }
}

In ronaldo$[i,j,z]$ I've stored the results of the conditional means at day $i$ of asset $j$ obtained in simulation $z$ (obtained by: i. inverse of CDF of copula realizations; ii. insert in GARCH formula; iii. compute conditional mean). [yes, Ronaldo is my favorite player, and I mean the brazilian Fenomeno]. I should get $p=1$ if any value stored in ronaldo is NaN, right? Well, $p=0$.

Is this possible? Is it possible that R produces a message which isn't verified by the results? Already at a graphical inspection everything seemed alright (I only had numerical values stored in ronaldo), and that little trick with $p$ verifies this. Am I justified in "moving on" and ignoring the message since I don't have NaNs?