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.

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


Like others have noted your book should have used $h \to 0^{+}$ (otherwise the step is invalid) and written an intermediate step (it must be written if the target audience is not an expert in calculus) as follows:
\begin{align}
L &= \lim_{h \to 0^{+}}\dfrac{\dfrac{h^{5}}{2h^{4}}}{\sqrt{h^{2} + h^{4}}}\notag\\
&= \lim_{h \to 0^{+}}\dfrac{\dfrac{h^{5}}{2h^{4}}}{h}\cdot\frac{h}{\sqrt{h^{2} + h^{4}}}\notag\\
&= \lim_{h \to 0^{+}}\frac{h^{5}}{2h^{5}}\cdot\frac{1}{\sqrt{1 + h^{2}}}\notag\\
&= \lim_{h \to 0^{+}}\frac{h^{5}}{2h^{5}}\cdot 1\notag\\
\end{align}
I wonder why books take such shortcuts and miss such non-trivial steps.
