Find function $ f(x) $ to ensure the limit has certain value If $\lim_{x\to1}$  $\frac{f(x)}{(x-1)(x-2)} = -3$ , then provide a possible function $y = f(x)$
*I don't understand what the question is asking me and how I should solve it. Can a possible function be $y = f(1)$? Or must I do something else to figure out the answer? I'd appreciate if anyone can help me out.
HINT
Let
$$f(x)=(x-1)g(x)$$
such that
$$\lim_{x\to 1}\frac{f(x)}{(x-1)(x-2)} = \lim_{x\to 1}\frac{(x-1)g(x)}{(x-1)(x-2)} =\lim_{x\to 1}\frac{g(x)}{x-2}=-3$$
Refer also to

Why are we allowed to cancel fractions in limits?
For making the limit to exist we gotta take the function as f(x)=(x-1)g(x).
So let's assume g(x) to be a polynomial function of degreer 1 i.e. g(x)=(x-n) where n is any real number.Therefore f(x)=(x-1)(x-n).
So just simply apply the limits and you will be left with (n-1).
Put it equal to -3 , you will get n = -2.
There you have your f(x)=(x-1)(x+2)