My school textbook has a derivation for rational functions and there is a step that just doesn't make any sense to me, can someone help me out?
Here's the excerpt from my textbook:
A function $f$ is said to be a rational function, if $f(x)=\dfrac{g(x)}{h(x)}$, where $g(x)$ and $h(x)$ are polynomials such that $h(x)\neq0$. Then $$\lim_{x\to a}f(x)=\lim_{x\to a}\frac{g(x)}{h(x)}=\frac{\lim\limits_{{x\to a}}g(x)}{\lim\limits_{x\to a}h(x)}=\frac{g(a)}{h(a)}$$
However, if $h(a)=0$, there are two scenarios - (i) when $g(a)\neq0$ and (ii) when $g(a)=0$. In the former case we say that the limit does not exist. In the latter case we can write $g(x)=(x-a)^kg_1(x)$, where $l$ is the maximum of powers of $(x-a)$ in $g(x)$. Similarly, $h(x)=(x-a)^lh_1(x)$ as $h(a)=0$. Now, if $k>l$, we have $$\lim_{x\to a}f(x)=\lim_{x\to a}\frac{g(x)}{h(x)}=\frac{\lim\limits_{{x\to a}}(x-a)^kg_1(x)}{\lim\limits_{x\to a}(x-a)^lh_2(x)}$$
If $g(a) = 0$, we can write $g(x) = (x – a)^k g_1(x)$, where $k$ is the maximum of powers of $(x – a)$ in $g(x)$ How?? Similarly, If $h(a) = 0,$ then $h(x) = (x – a)^l h_1(x)$
What function is is $g_1(x)$ where did $k$ and $(x-a)$ come from?
You can download the whole chapter of my textbook on Limits & derivatives from,
To start from the beginning, the function $f(x)=\frac{g(x)}{h(x)}$ is a polynomial $g(x)$ divided by another polynomial $h(x)$. ($h(x)\ne 0.\text{Why?}$) To evaluate $\lim_{x\to a}f(x)=\lim_{x\to a}\frac{g(x)}{h(x)}$, we are only concerned with the case $h(x)=0$. Otherwise, the value is a rational number.
If $h(a)=0$, an indeterminate form $\frac{0}{0}$ arises if $g(a)=0$ or else the limit would not exist.
Since $g(a)=h(a)=0$, it is evident that $(x-a)$ is a factor of both denominator and numerator.(Remainder Theorem)
Hence, $g(x)=(x-a)^kg_1(x)$. We are not sure how many times $x-a$ is contained in the polynomial. Where $g_1(x)$ is the new polynomial obtained after dividing by $(x-a)$ k times.[Factor Theorem]
Similarly, $h(x)$ can be written as $(x-a)^th_1(x)$ since $x-a$ is definitely a factor of $h(x)$ since $h(a)=0$
For quick example, consider the function $f(x)=\frac{x^3-5x^2+8x-4}{x^3-4x^2+5x-2}$
Try and evaluate $\lim_{x\to 1}f(x) \text{ and } \lim_{x\to 2}f(x)$. They will both be of indeterminate form $\frac{0}{0}$.
As you should verify, $g(1)=0, g(2)=0, h(1)=0, h(2)=0$ and $\therefore f(x)$ can be rewritten as $\frac{(x-2)^2(x-1)}{(x-1)^2(x-2)}$