Where is the flaw in my calculation of $\lim\limits_{x \to 0} \frac{x^2}{1-\cos(x)}$

Question

The limit at 0 may be correctly calculated this way: $$ \lim\limits_{x \to 0} \frac{x^2}{1-\cos(x)} = \lim\limits_{x \to 0} \frac{x^2}{1-\cos(x)}\cdot\frac{1+\cos(x)}{1+\cos(x)}=\lim\limits_{x \to 0} \frac{x^2(1+\cos(x))}{\sin^2(x)}=\\\lim\limits_{x \to 0} \left(\frac{x^2}{\sin^2(x)}\cdot(1+\cos(x)\right)=\lim\limits_{x \to 0} \frac{x^2}{\sin^2(x)}\cdot\lim\limits_{x \to 0}(1+\cos(x)=1\cdot2=2 $$

Where is the flaw in the following? $$\lim\limits_{x \to 0} \frac{x^2}{1-\cos(x)} = \lim\limits_{x \to 0} \frac{x}{1-\cos(x)}\cdot\lim\limits_{x \to 0}\,x$$

Thank you.

Answer 1

This step

$$\lim\limits_{x \to 0} \left(\frac{x^2}{\sin^2(x)}\cdot(1+\cos(x)\right)=\lim\limits_{x \to 0} \frac{x^2}{\sin^2(x)}\cdot\lim\limits_{x \to 0}\left(1+\cos(x)\right)$$

is allowed because both limits exist finite, but here

$$\lim\limits_{x \to 0} \frac{x^2}{1-\cos(x)} = \lim\limits_{x \to 0} \frac{x}{1-\cos(x)}\cdot\lim\limits_{x \to 0} x$$

as noticed in the comments, the first limit diverges while the second one is equal to zero (i.e. it is an indeterminate form $\infty \cdot 0$), therefore the step is not allowed, as dicussed in detail in the following answer.

Answer 2

Bit of a late response and this may not be the intended method but you can also use l'Hôpital's rule twice for this:

$$\lim_{x\to 0}\frac{x^2}{1-\cos(x)}=\lim_{x\to 0}\frac{2x}{\sin(x)}=\lim_{x\to 0}\frac{2}{\cos(x)}=2$$