Bizarre limit, neat answer

As Brian Charlesworth observed in his solution to an earlier problem , $\sum_{k=1}^{\infty}\frac{e^{ikx}}{k}=-\ln(1-e^{ix})$ for $0<x<\pi$ . Taking real parts and recalling that $Re(\ln(z))=\ln|z|$ , we find that $\sum_{k=1}^{\infty}\frac{\cos(kx)}{k} =-\ln|1-e^{ix}|=-\ln\left(2\sin(x/2)\right)$ To prove the last equation geometrically, consider the triangle in the Argand plane with its vertices at 0, 1, and $e^{ix}.$ Finally $\lim_{x\to{0^+}}\sum_{k=1}^{\infty}\frac{\cos(kx)}{k\ln(x)} =-\lim_{x\to{0^+}}\frac{\ln(2\sin(x/2))}{\ln(x)}=-\lim_{x\to{0^+}}\frac{(x/2)\cos(x/2)}{\sin(x/2)}=-1$ by Bernoulli's rule (often mistakenly referred to as L'Hôpital's rule).