Show that : $\lim\limits_{h\to 0}\frac {h^5} {2h^4} \frac1{\sqrt{h^2 + h^4}} = \lim\limits_{h \to 0}\frac {h^5} {2h^5}$ In my textbook, I found the following step, but I don't understand how the author gets there.

$$\lim_{h\to 0} {{\frac {h^5} {2h^4} \over \sqrt{h^2 + h^4}}} =  \lim_{h \to 0}\frac {h^5} {2h^5}$$


The comment by Claude explains the step of that author but only for $\;h>0\;$, but the step is wrong for negative values of $\;h\;$ :
$$\frac{\frac{h^5}{2h^4}}{\sqrt{h^2+h^4}}=\frac h{2|h|\sqrt{1+h^2}}=\begin{cases} \cfrac1{2\sqrt{1+h^2}}, &h>0\xrightarrow[h\to0^-]{}-\cfrac12\\{}\\-\cfrac1{2\sqrt{1+h^2}},&h<0\xrightarrow[h\to0^+]{}\cfrac12\end{cases}$$
and thus the limit doesn't exist.
