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.
 A: 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

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*Why are we allowed to cancel fractions in limits?
A: Suppose you wanted to get rid of the discontinuity at $x=1$. Create a function $f(x)$ that removes this discontinuity, that is, $f(x)=(x-1)g(x)$.
A: You want to remove the discontinuity produced by $ x-1 $ in the denominator so $ f(x) $ should have $ (x-1) $ as one of the factors
With this knowledge, let's say $ f(x) = (x-1) \cdot a $
\begin{align}
\frac{(x-1)\cdot a}{(x-1)(x-2)} = &-3 \\
\frac{a}{(x-2)} = &-3 \\
a = & -3(x-2) \\
a = & -3x + 6 \\
\end{align}
So one option could be $ f(x) = (x-1)(-3x+6) $ or expanded: $ -3x^2 + 9x - 6 $
As the $ x\to 1 $ the denominator, once the discontinuity is removed, tends to $ -1 $ so any numerator that includes $ x-1 $ and another factor that equals to $ 3 $ will be a solution.
Another option: $  f(x) = 3(x-1) = 3x-3 $
A: Well, if the numerator of the limit is $-3$ times the denominator, we are done. Hence, it suffices to choose $f(x)=-3(x-1)(x-2)$.
A: 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)
