Limit with absolute-value function on denominator I am trying to evaluate the following limit:
$$\lim_{x \to 0}  \frac{m^2x^3}{\lvert 3mx^2\rvert^{3/2}}$$
where $m \in \Bbb{R}$.
If we allow $m$ to vary, that renders the limit to be nonexistent, am I correct?

Since $\vert x^2\vert^{3/2}=x^2\vert x\vert$ the expression simplifies to
\begin{equation}
\lim_{x\to0}\frac{\sqrt{\vert m\vert}}{3}\cdot\frac{x}{\vert x\vert}
\end{equation}
so the left and right limits differ. Thus the limit does not exits.
