Let $f(x)=x{\times}(-1)^{\left \lfloor \frac1x \right \rfloor}$. Calculate its limit at $x=0$. According to me the limit doesn't exist because if I take log on both sides of the equation, I get:- $$\ln f(x) = \ln x+\left \lfloor \frac1x \right \rfloor {\times} \ln(-1)$$
Here $\ln(-1)$ doesn't exist and hence no limit should exist.