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if :

$f(x),g(x).h(x),k(x)⇒$ Polynomial

and:

$$\lim_{ x \to a }(\frac{f(x)}{g(x)}-\frac{h(x)}{k(x)})→∞-∞$$

then:

$$\lim_{ x \to a }(\frac{f(x)}{g(x)}-\frac{h(x)}{k(x)})=?$$

such as :

$$\lim_{ x \to 1 }(\frac{1}{x-1}-\frac{3x-1}{x^2-1})=-1$$

thank you very much !!

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    $f,g,h,...k?$ Oh come on, why you gotta be hating on $i(x)$? – 2017-01-12
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    @SimpleArt ,General answer ! :) is possible ؟؟ – 2017-01-12

2 Answers 2

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We have

$$\frac{f(x)}{g(x)}-\frac{h(x)}{k(x)}=\frac{f(x)k(x)-g(x)h(x)}{g(x)k(x)}$$

Now, there isn't a universal solution, but the general idea is that for you to have $\infty-\infty$, then $g(x)$ and $k(x)$ both end up equaling zero, or

$$g(x)k(x)=(x-a)^2P(x)$$

For the limit to exist, we must then have

$$f(x)k(x)-g(x)h(x)=(x-a)^2Q(x)$$

to avoid division by $0$ again, and thus,

$$\lim_{x\to a}\frac{f(x)}{g(x)}-\frac{h(x)}{k(x)}=\frac{Q(a)}{P(a)}$$

If there does not exist $Q(x)$, then the limit does not exist.

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    Have you always$f(x)k(x)-g(x)h(x)=(x-a)^2Q(x) , g(x)k(x)=(x-a)^2P(x)$?? – 2017-01-12
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    @Almot1960 Not always will there be a $Q(x)$, as I point out at the last sentence. But there will always be a $P(x)$. And I am assuming that "polynomials" have positive whole number exponents (for my sanity) – 2017-01-12
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    You're welcomed very much @Almot1960 – 2017-01-13
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$\lim_\limits{ x \to a }(\frac{f(x)}{g(x)}-\frac{h(x)}{k(x)})=\lim_\limits{ x \to a }(\frac{fk - hg}{gk})$

If $\lim_\limits{ x \to a }\frac{f(x)}{g(x)} = \infty$ we know that $a$ is a root of $g(x)$ And by similar logic we know that $a$ is a root of $k(x)$.

if $\lim_\limits{ x \to a }(\frac{fk - hg}{gk}) = L$ then $a$ is a root of multiplicity of $fk - hg$