This rule only applies to these two indeterminate forms: $\frac 0 0 \text { or } \frac {\pm\infty} {\pm \infty}$
$$ \text {if } \lim_{x→c} \frac {f(x)} {g(x)} ={\frac 0 0} \text { or } \frac {\pm\infty} {\pm{\infty}}, \text { then } \lim_{x→c}{\frac {f(x)} {g(x)}}=\lim_{x→c}{\frac {f'(x)} {g'(x)}} $$
You can also use this rule with multiple prime functions for…
$$ \lim_{x→c}{\frac {f(x)} {g(x)}}=\lim_{x→c}{\frac {f'(x)} {g'(x)}}=\lim_{x→c}{\frac {f''(x)} {g''(x)}}=\lim_{x→c}{\frac {f'''(x)} {g'''(x)}}...... $$
$$ \lim_{x→a} \frac {f'(x)} {g'(x)}= \frac {\lim_{x→a} {f'(x)}} {\lim_{x→a} {g'(x)}} $$
$$ \frac {\lim_{x→a} {f'(x)}} {\lim_{x→a} {g'(x)}}=\frac { {\lim_{x→a} \frac {f(x)-f(a)} {x-a}}}{ {\lim_{x→a} \frac {g(x)-g(a)} {x-a}}}=\frac { {\lim_{x→a} {f(x)-f(a)}}}{ {\lim_{x→a} {g(x)-g(a)}}} $$
Formula/Definiton of Derivative
$$ \frac { {\lim_{x→a} {f(x)-f(a)}}}{ {\lim_{x→a} {g(x)-g(a)}}}= \frac { {\lim_{x→a} {f(x)-[0]}}}{ {\lim_{x→a} {g(x)-[0]}}} $$
Remember the indeterminate form is $\frac 0 0 \text { or } \frac {\pm\infty} {\pm \infty}$
In this case sake, use the 0/0
$$ \frac { {\lim_{x→a} {f(x)}}}{ {\lim_{x→a} {g(x)}}}= \lim_{x→a}\frac {f(x)} {g(x)} $$
Therefore……
The rule exist.