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 step is based on the “product/quotient rule” of limits. Suppose you have two functions $f(x)$ and $g(x)$ defined in a neighborhood of $a$ and such that $\lim_{x\to a}g(x)=1$. Then you can say that
$$
\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)
$$
in the sense that the either both limit exist or both limits don’t exist; in case one exists, they are equal.
Indeed, suppose $\lim_{x\to a}f(x)$ exists. By the product rule,
$$
\lim_{x\to a}f(x)=\Bigl(\lim_{x\to a}f(x)\Bigr)\cdot 1=
\Bigl(\lim_{x\to a}f(x)\Bigr)\cdot\Bigl(\lim_{x\to a}g(x)\Bigr)=
\lim_{x\to a}f(x)g(x)
$$
Conversely, if $\lim_{x\to a}f(x)g(x)$ exists, we have
$$
\lim_{x\to a}f(x)g(x)=
\Bigl(\lim_{x\to a}f(x)g(x)\Bigr)\cdot1=
\Bigl(\lim_{x\to a}f(x)g(x)\Bigr)\cdot
\Bigl(\lim_{x\to a}\frac{1}{g(x)}\Bigr)=
\lim_{x\to a}f(x)g(x)\frac{1}{g(x)}=
\lim_{x\to a}f(x)
$$
Limits can also be one-sided, with the same argument.
In your case you can consider
$$
g(h)=\frac{\sqrt{h^2+h^4}}{h}
$$
and observe that
$$
\lim_{h\to0^+}g(h)=1
$$
Thus, by the argument above
$$
\lim_{h\to0^+}\frac{\frac{h^5}{2h^4}}{\sqrt{h^2 + h^4}}=
\lim_{h\to0^+}\frac{\frac{h^5}{2h^4}}{\sqrt{h^2 + h^4}}
  \underbrace{\frac{\sqrt{h^2+h^4}}{h}}_{\text{has limit $1$}}=
\lim_{h\to0^+}\frac{h^5}{2h^4}\frac{1}{h}
$$
Unfortunately, the textbook’s argument is wrong, because the given limit doesn’t exist: the limit from the right and from the left are different. The error is due to the false assumption that $\sqrt{h^2}=h$, a typical beginner’s mistake.
