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;
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} = 0$$

Too long for comments.
Your work is fine but I think that there is a faster solution.
When I see a problem such as $\lim_{x \to a} f(x)$, my first reaction is to let $x=y+a$ and consider $\lim_{y \to 0} g(y)$.
If we do it for your problem, we have
$$\lim_{x \to 1} \cos(\pi  x) \, \sqrt{\frac{(x-1)^2}{(x^2-1)}} \, \frac{1-x}{x^2+x-1}=\lim_{y \to 0}\cos (\pi  y)\frac{y  }{y^2+3 y+1}\sqrt{\frac{y}{y+2}}$$ which gives $1\times 0\times 0=0$
