How to calculate $\lim_{x\to 1} \frac{(x^2-3x+2)}{(x-1)(x^3-1)}$?

How to calculate $\lim_{x\to 1} \frac{(x^2-3x+2)}{(x-1)(x^3-1)}$?

I tried: $$\lim_{x\to 1} \frac{(x-1)(x-2)}{(x-1)(x^3-1)}$$ $$\lim_{x\to 1} \frac{x-2}{x^3-1} $$

On this step i don't know what to do. I tried to expand the denominator using a³ - b³ = (a-b)(a²+b²+ab) it didn't work. I also try to divide both numerator and denominator by x but that also not working.

I checked from desmos and found the limit doesn't exists but how to show that ?

Notice that by writing

$$f(x) = \frac{x-2}{x^2 + x + 1}$$

you have that

$$\frac{x-2}{x^3-1} = \frac{x-2}{x^2 + x + 1} \cdot \frac{1}{x-1} = \frac{f(x)}{x-1}$$

Notice, moreover, that as $x \to 1$, $f(x) \to -1/3$. Meanwhile, depending on which direction you approach from, $1/(x-1) \to \pm \infty$. The end result is clear: your one-sided limits go to $+\infty$ and $-\infty$ and hence the limit overall does not exist.