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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.

Jon
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Panda14
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5 Answers5

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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

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

C Squared
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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 $

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

Zuy
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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)

parth agarwal
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