How to fix two singularities

For $\displaystyle\frac{f(x)}{g(x)}$ , we know that since both function is undefined at $x =1$ , then we multiply the fraction with $\displaystyle\frac{x-1}{x-1}$ , which gives

$\frac{f(x)}{g(x)} = \frac{1}{\frac{3}{x-2}} = \frac{x-2}{3}\Rightarrow \lim_{x\rightarrow1}\frac{f(x)}{g(x)} = -\frac{1}{3}$

f(x)/g(x)=x^2-3x+2/3(x-1) x^2-3x+2=(x-1)(x-2) =>f(x)/g(x)=(x-1)(x-2)/3(x-1) =(x-2)/3 lim x->1 f(x)/g(x)=1/3 lim x->1 (x-2) =1/3*-1 =-1/3

First: $\frac{\frac{1}{x-1}}{\frac{3}{x^2 - 3x + 2}} = \frac{x^2 - 3x + 2}{3x - 3}$

Since $\underset{x \rightarrow 1}{\lim} x^2 - 3x + 2 = 0 \wedge \underset{x \rightarrow 1}{\lim} 3x - 3 = 0$ you can also use the L'Hôpitals rule to get $\underset{x \rightarrow 1}{\lim} \frac{x^2 - 3x + 2}{3x - 3} = \underset{x \rightarrow 1}{\lim} \frac{2x-3}{3} = \frac{-1}{3}$

That was so easy. Easy money easy life.

This is one of those factor, cancel, plug it in sort of limit problems. g(x) = (x-1)(x-2). Eliminate the common factor of (x-1) in f and g and we are left with (x-2)/3 no problem. Why can we "plug it in"? Because (x-2)/3 is defined, continuous at 1, and equal to f/g everywhere but 1.

It's too simple,☺😊.. however it's good to memorize.