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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.

Our
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    Can the one who voted down give a reason please ? – Our Jul 27 '16 at 08:36
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    because your problem lost some details about what you have done. – Zau Jul 27 '16 at 10:10
  • It is because I couldn't even move a pen, since couldn't figure out what the limits are.So ? – Our Jul 27 '16 at 13:36
  • @Op No that's not a sound reason because I can figure out the limit by my brain and my major is not math but computer science. It only turn out that you are unfamiliar with limit. What's more, if you allow me to allude more, I can totally conclude that the problem is copied even it may be a problem which is your homework or quest for exam. – Zau Jul 27 '16 at 14:10
  • So, you concluded that I'm a stereotype math student just from my not being able to construct at least a basic idea ? – Our Jul 27 '16 at 14:55
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    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? – Martin Sleziak Jul 30 '16 at 08:38
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    I edited, in the book the equation was without "=0"; therefore, I directly wrote like that. – Our Jul 30 '16 at 09:19
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    Since you write that it was written in the book, including the name of the book would, in my opinion, count as adding context. – Martin Sleziak Jul 30 '16 at 09:28

4 Answers4

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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.

Now,

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

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.

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rubik
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  • what do you mean by "from the definition of limit and the intermediate value theorem" ? – Our Jul 27 '16 at 08:25
  • @Leth Added a little explanation. – rubik Jul 27 '16 at 08:26
  • I still couldn't understand how the function is continuous on [a,b] ? – Our Jul 27 '16 at 08:29
  • @Leth That's not a problem since it's continuous on $\mathbb R \setminus {1,2}$ (i.e. its domain). And $(1,2) \subset \operatorname{dom} f$. What is exactly that isn't clear for you? – rubik Jul 27 '16 at 08:30
  • Ok, it is continuous over its domain except {1,2} but this exception should contradict with closed interval obligation. – Our Jul 27 '16 at 08:33
  • which a and b are you talking about ? a> 1 and b<2 ? – Our Jul 27 '16 at 08:33
  • @Leth $a$ and $b$ are just general variables that are in IVT's statement. Here, the definition of limit guarantees us that there are two constants $\delta,\delta'$ such that if we are near enough to $1$ (where "enough" is defined by $\delta$) then $f$ is positive and greater than whichever $\varepsilon$ we choose. Similarly for the second limit. – rubik Jul 27 '16 at 08:40
  • Ok, after examine what you have written, I get it but there is some problem like, this approach is taught as not valid.I mean if f is continuous on (c,d) then you cannot apply IVT – Our Jul 27 '16 at 08:41
  • @Leth Oh no, $f$ is continuous on $[1 + \delta, 2 - \delta']$, not an open interval! – rubik Jul 27 '16 at 08:43
  • No, I mean if f is continuos on some open interval, it is obvious that there is some closed interval in that open interval which the function is continuous. – Our Jul 27 '16 at 08:45
  • @Leth Yes, that follows easily from the definition of continuity. A function $g$ is continuous on an interval if it's continuous in every point of the interval. If you take a subset of said interval, the property is still satisfied. Why do you say it's not valid? – rubik Jul 27 '16 at 08:47
  • I'm not saying, it is told me by the instructor of the Calculus course. – Our Jul 27 '16 at 09:44
  • By the way,Doesn't the fact that $\lim_{x\to 1^+} f(x)$ and $\lim_{x\to 2^-} f(x)$ are not exist cause any trouble ? – Our Jul 29 '16 at 07:43
  • @Leth But they do exist. They're, respectively, $+\infty$ and $-\infty$. If you meant to ask instead "does the fact the the function doesn't exist in $1$ and $2$ cause problems?", then the answer is that when we take limits, we are not at all interested in the value the function attains in that point (if it even has a value), but in what happens in a neighborhood of that point. In this case, we are interested in the behaviour in a right neighborhood of $1$ and in a left neighborhood of $2$. – rubik Jul 29 '16 at 09:57
  • No,it is wrong.If limit of a function goes to infinity at some point, then the limit doesn't exist, which it is the case. – Our Jul 29 '16 at 17:16
  • @Leth That depends on the definition of limit you're using. But consider that if you treat infinite limits like they don't exist then you will not be able to distinguish them from other limits that do not exist according to the definition I'm using here (and which is the most used one). An example of this is $\lim\limits_{x \to +\infty} \sin x$. It's much better to say that the limits in my answer are not finite, while this last one does not exist. Clearly they are not the same. – rubik Jul 29 '16 at 19:01
  • Yes, they are not the same but, as far as I know, the fact that limit doesn't exist when the limit goes to infinity is still valid for both limits.Could you support your claim by giving some reference to your statement ? – Our Jul 30 '16 at 04:21
  • @Leth This site contains all the definitions. Again, saying that a limit does not exist finite and that it does not exist are two very different things. Putting it all together does not allow you to distinguish fundamentally different behaviors. – rubik Jul 30 '16 at 06:28
  • I didn't get it. – Our Jul 30 '16 at 07:23
  • @Leth I encourage you to ask a new question to resolve your confusion. – rubik Jul 30 '16 at 07:24
  • with also citing your answer ? – Our Jul 30 '16 at 07:25
  • @Leth If you want, yes. – rubik Jul 30 '16 at 07:26
  • Thanks to Zack Ni,who reported my all question with missing context, I 'm forbidden from asking question :) – Our Jul 30 '16 at 07:28
  • @Leth That's weird. You should go on meta.math.stackexchange.com to resolve the issue. – rubik Jul 30 '16 at 07:30
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    @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 http://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. – user21820 Jul 31 '16 at 06:35
  • @user21820 Ok, if we skip whether the limit exist or not, which I'm convinced there is, in order to use IVT, the function must be continuous on a closed interval.@rubik showed that there is closed interval but how can we say that the function is continuous at the end points of that interval from the definition of continuity ? – Our Jul 31 '16 at 07:46
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    @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. – rubik Jul 31 '16 at 08:05
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    @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.) – user21820 Jul 31 '16 at 08:30
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In the interval $(1,2)$ you may freely multiply by $(x-1)(x-2)$ and

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

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

Clearly,

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

  • 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. – Our Jul 30 '16 at 13:52
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    @Leth: was this ever explained in your post ? Sorry, I can't read your mind. By the way, my answer is a rigorous proof. –  Aug 12 '16 at 06:24
  • 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. – Our Aug 12 '16 at 06:38
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    @Leth: what frustrates me is that I still don't know what you are after, as you didn't state it. –  Aug 12 '16 at 06:41
  • I was after a solution including basic theorems of calculus. – Our Aug 12 '16 at 06:44
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Hint

Define (and fill in details)

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

so

$$\lim_{x\to1^+}f(x)=(-A)\lim_{x\to1^+}\frac1{(x-1)(x-2)}=\infty$$

$$\lim_{x\to2^-}f(x)=B\lim_{x\to2^-}\frac1{(x-1)(x-2)}=-\infty$$

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

DonAntonio
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  • But in order to use IVT, the function must be continuous at some closed interval ? – Our Jul 27 '16 at 08:22
  • @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 $;[1+\epsilon,,2-\delta];$ .... – DonAntonio Jul 27 '16 at 08:28
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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 $f(\frac{2A+B}{A+B}) = 0$.