Find the limiting value of lim x->2 [2x-2]/[x-1]

$$
\lim_{x \to 2} \left\{ \frac{2x-2}{x-1} \right\} = \frac{2(2)-2}{2-1}
$$

$$
= \frac{4-2}{1} = \frac{2}{1} = 2
$$
The limiting value is $\boxed{2}$.

$$
x^2-5x+6 = (x-2)(x-3)
$$

$$
\lim_{x \to 3} \frac{x-3}{(x-2)(x-3)} = \lim_{x \to 3} \frac{1}{x-2}
$$

$$
= \frac{1}{3-2} = \frac{1}{1} = 1
$$
The limit is $\boxed{1}$.

$$
\lim_{x \to 0} \frac{x(2x^2-3x+4)}{x(5x^2+2x-3)}
$$

$$
= \lim_{x \to 0} \frac{2x^2-3x+4}{5x^2+2x-3}
$$

$$
= \frac{2(0)^2-3(0)+4}{5(0)^2+2(0)-3} = \frac{0-0+4}{0+0-3} = \frac{4}{-3}
$$
The limit is $\boxed*{-\frac{4}{3}*}$.

$$
(3x-4)(2x+3) = 6x^2+9x-8x-12 = 6x^2+x-12
$$
$$
(5x-2)(x+6) = 5x^2+30x-2x-12 = 5x^2+28x-12
$$
Numerator:
$$
(6x^2+x-12) - (5x^2+28x-12) = 6x^2+x-12-5x^2-28x+12
$$
$$
= (6x^2-5x^2) + (x-28x) + (-12+12) = x^2-27x
$$

$$
x^2-27x = x(x-27)
$$

$$
\lim_{x \to 0} \frac{x(x-27)}{x(x^2-9)} = \lim_{x \to 0} \frac{x-27}{x^2-9}
$$

$$
= \frac{0-27}{0^2-9} = \frac{-27}{-9} = 3
$$
The limit is $\boxed{3}$.