# 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.

• $$f(x)=-3(x-1)(x-2)$$ Commented Aug 7, 2020 at 21:21
• The question is asking you for a functional form of $f(x)$, for example can you compute the limit with $f(x)$ as the function mentioned by @PeterForeman above? Commented Aug 7, 2020 at 21:22

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

Suppose you wanted to get rid of the discontinuity at $$x=1$$. Create a function $$f(x)$$ that removes this discontinuity, that is, $$f(x)=(x-1)g(x)$$.

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: $$f(x) = 3(x-1) = 3x-3$$

Well, if the numerator of the limit is $$-3$$ times the denominator, we are done. Hence, it suffices to choose $$f(x)=-3(x-1)(x-2)$$.