If $A$ and $B$ are positive constants, show that $\frac{A}{x-1} + \frac{B}{x-2}$ has a solution on $(1,2)$

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 $B = 1$):

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

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.

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 $\;(1,2)\;$