Calculate the following limit:
$$\displaystyle\lim_{x \to 2}{\dfrac{x^2-3x+2}{x-2}}$$
Development:
$$\displaystyle\lim_{x \to 2}{\dfrac{x^2-3x+2}{x-2}}=\dfrac{4-6+2}{0}=\dfrac{0}{0}$$
Since $$2$$ cancels the polynomial of the numerator, we factor it:
$$x^2-3x+2=(x-1)\cdot(x-2)$$
$$\displaystyle\lim_{x \to 2}{\dfrac{(x-1)(x-2)}{(x-2)}}=\lim_{x \to 2}{(x-1)}=2-1=1$$
Solution:
$$1$$