I have problem which I couldn't figure out how to solve; If $A$ and $B$ are positive constants, show that $$0=\frac{A}{x-1} + \frac{B}{x-2}$$ has a solution on the open interval $(1,2)$.

If you support your answers with rigorous proof, I appreciate that.

What I thought was taking interval roughly close to the end points from the inside i.e $[1.1,1.9]$, but then it wouldn't be rigorous solution to this problem. After that, I totally stuck since I couldn't determine to closed interval, which prevented me from using any useful theorem.

Note: The problem is taken from G.Simmons Calculus with Analytic Geometry 2nd.

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    $\begingroup$ Can the one who voted down give a reason please ? $\endgroup$
    – Our
    Jul 27, 2016 at 8:36
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    $\begingroup$ because your problem lost some details about what you have done. $\endgroup$
    – Zau
    Jul 27, 2016 at 10:10
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    $\begingroup$ I think that "unclear what you are asking would be more appropriate. What do you mean by $\frac A{x-1}+\frac B{x-2}$ has a solution? Did you want to write that it has a zero or a root? Or to ask whether $\frac A{x-1}+\frac B{x-2}=0$ has a solution? $\endgroup$ Jul 30, 2016 at 8:38
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    $\begingroup$ I edited, in the book the equation was without "=0"; therefore, I directly wrote like that. $\endgroup$
    – Our
    Jul 30, 2016 at 9:19
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    $\begingroup$ Since you write that it was written in the book, including the name of the book would, in my opinion, count as adding context. $\endgroup$ Jul 30, 2016 at 9:28

4 Answers 4


Let $$f(x) = \frac{A}{x - 1} + \frac{B}{x - 2}.$$

Observe that $f$ is defined on $\mathbb R \setminus \{1,2\}$ and it's continuous since it's a sum of continuous functions.


  • $\lim\limits_{x \to 1^+} f(x) = +\infty$
  • $\lim\limits_{x \to 2^-} f(x) = -\infty$

Therefore, from the definition of limit and the intermediate value theorem, it follows that $f$ has a root on $(1, 2)$.

It is important to note that the result follows from the definition of limit as well, because the intermediate value theorem requires $f$ to be continuous over a compact $[a, b]$. Indeed, we have $$\forall \varepsilon > 0, \exists \delta \text{ such that } x - 1 < \delta \implies f(x) > \varepsilon$$ And similarly, from the second limit, $$\forall \varepsilon' > 0, \exists \delta' \text{ such that } 2 - x < \delta' \implies f(x) < -\varepsilon'$$ So it's possible to choose appropriate constants $\delta,\delta'$ such that $f$ satisfies the conditions of the intermediate value theorem on $[1 + \delta, 2 - \delta']$ and $f(1 + \delta') > 0$ and $f(2 - \delta) < 0$.

The graph may make the reasoning clearer ($A = 2$ and $B = 1$): graph

You may also be interested in this very similar question, which has a similar (albeit a bit more involved) solution.

  • $\begingroup$ what do you mean by "from the definition of limit and the intermediate value theorem" ? $\endgroup$
    – Our
    Jul 27, 2016 at 8:25
  • $\begingroup$ @Leth Added a little explanation. $\endgroup$
    – rubik
    Jul 27, 2016 at 8:26
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    $\begingroup$ @Leth: There is good reason for using your book's definition of existence of limits, because the rules for addition and multiplication of limits are simpler (see math.stackexchange.com/a/1782096/21820). However, although rubik used a different definition of limit existence in his answer, his conclusion is valid because it does not mention limit existence anywhere. Go back to the rigorous definition of $\lim_{x \to 1^+} f(x) = \infty$. It tells you that there is some $a > 1$ such that ( $f(x) > 1000$ for every $x \in (1,a]$ ). Use $a$ for the left point for IVT. Same on the right. $\endgroup$
    – user21820
    Jul 31, 2016 at 6:35
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    $\begingroup$ @Leth That was discussed before here. If you already know/proved that $f$ is continuous (i.e. in its domain), you are done. Otherwise just use the definition, it's a fairly simple application of it. $\endgroup$
    – rubik
    Jul 31, 2016 at 8:05
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    $\begingroup$ @Leth: Any function that is continuous on some domain is continuous on any subdomain just by definition of continuity. (The reverse might not be true.) $\endgroup$
    – user21820
    Jul 31, 2016 at 8:30

In the interval $(1,2)$ you may freely multiply by $(x-1)(x-2)$ and

$$A(x-2)+B(x-1)=0$$ or

$$(A+B)x=2A+B,$$ $$x=\frac{2A+B}{A+B}=1+\frac A{A+B}.$$


$$0<\frac A{A+B}<1,$$ which substantiates the claim.

  • $\begingroup$ Yes, it is clearly one of the solutions of the problem, thanks but as I'm not interested in solving with algebraic manipulation, it doesn't help me understating the topic better. $\endgroup$
    – Our
    Jul 30, 2016 at 13:52
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    $\begingroup$ @Leth: was this ever explained in your post ? Sorry, I can't read your mind. By the way, my answer is a rigorous proof. $\endgroup$
    – user65203
    Aug 12, 2016 at 6:24
  • $\begingroup$ I didn't mean to be rude, I just did an explanation for not accepting as an answer.By the way, I never said it is not a rigorous proof, I just said the method is a manipulation. $\endgroup$
    – Our
    Aug 12, 2016 at 6:38
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    $\begingroup$ @Leth: what frustrates me is that I still don't know what you are after, as you didn't state it. $\endgroup$
    – user65203
    Aug 12, 2016 at 6:41
  • $\begingroup$ I was after a solution including basic theorems of calculus. $\endgroup$
    – Our
    Aug 12, 2016 at 6:44


Define (and fill in details)

$$f(x):=\frac{A}{x-1} + \frac{B}{x-2}=\frac{(A+B)x-2A-B}{(x-1)(x-2)}$$




and now use the IVT and the fact $\;f(x)\;$ is continuous at $\;(1,2)\;$

  • $\begingroup$ But in order to use IVT, the function must be continuous at some closed interval ? $\endgroup$
    – Our
    Jul 27, 2016 at 8:22
  • $\begingroup$ @Leth Of course. That's part of the "details". For example, we know there exists $\;\epsilon>0\;$ s.t. $\;f(1+\epsilon)=2\;$ , and also exists a $\;\delta>0\;$ such that $\;f(2-\delta)=-1\;$ . Well, now focus on $\;[1+\epsilon,\,2-\delta]\;$ .... $\endgroup$
    – DonAntonio
    Jul 27, 2016 at 8:28

To add on DonAntonio's Answer:

$$f(x) = \frac{A}{x-1} + \frac{B}{x-2} = \frac{(A+B)x - 2A - B}{(x-1)(x-2)}.$$

Now, $1 < \frac{2A+B}{A+B} < 2$, since $A,B > 0$. Moreover, one calculates $f(\frac{2A+B}{A+B}) = 0$.


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