Note that $\left\lfloor \dfrac1x \right\rfloor$ denotes the greatest integer less than or equals $\dfrac1x$. Hence, $(-1)^{\left\lfloor \dfrac1x \right\rfloor}$ makes sense since the power is always an integer. All you need for this proof is that $(-1)^{\left\lfloor \dfrac1x \right\rfloor}$ is either $1$ or $-1$. Hence, we have that $-x \leq x \times (-1)^{\left\lfloor \dfrac1x \right\rfloor} \leq x$ Hence, as $x \to 0$, we have that $\lim_{x \to 0}-x \leq \lim_{x \to 0} x \times (-1)^{\left\lfloor \dfrac1x \right\rfloor} \leq \lim_{x \to 0} x$ Hence, $\lim_{x \to 0} x \times (-1)^{\left\lfloor \dfrac1x \right\rfloor} = 0$
EDIT
Note that $\log(a^b) = b \log(a)$ is valid only when $a>0$ and $x \in \mathbb{R}$. Hence, it is incorrect to write $\log((-1)^b) = b \log(-1)$.