Find $\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$ (I need a review of my resolution please :) ) Find the limit:
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
This is what I have, i'm not sure about my answer (I'm just learning limits).
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \lim_{x \to 1} \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} $$
And I know that cos is a bounded limit, also its $L=-1$.
The other limit:
$$ \lim_{x \to 1} \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} $$
$$ \lim_{x \to 1} \sqrt{\frac{(x-1) \cdot (x-1)}{(x-1)\cdot (x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
Simplifying:
$$ \lim_{x \to 1} \sqrt{\frac{(x-1) \cdot (x-1)}{(x-1)\cdot (x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
$$ \lim_{x \to 1} \sqrt{\frac{(x-1)}{(x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
Evaluating:
$$ \lim_{x \to 1} \sqrt{\frac{(1-1)}{(1+1)}} \cdot \frac{1-1}{1^2+1-1} $$
$$ = \sqrt{0} \cdot \frac{0}{1} $$
$$\fbox {= 0}$$
So since I have a bounded limit and the other limit function is zero, the whole limit is $0$.
So;
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} = 0$$

Your answer is correct.
The limit can be split into parts if each of them exists as follows
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
$$=\lim_{x \to 1} \cos(\pi \cdot x) \cdot \lim_{x \to 1}\sqrt{\frac{x-1}{x+1}} \cdot \lim_{x \to 1}\frac{1-x}{x^2+x-1}$$
$$=(-1)\cdot 0\cdot 0=\color{blue}{0}$$
