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 :) )

Question

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;

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

Answer 1

Your answer is correct.

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

$$=(-1)\cdot 0\cdot 0=\color{blue}{0}$$

Answer 2

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 $1\times 0\times 0=0$