Given: $P(x)$ is a polynomial such that the following apply:
I) $\lim_\limits{x \to 1} \dfrac{P(x)}{(x-1)^3} = 6$
II). $\lim_\limits{x \to -2} \dfrac{P(x)}{x^2+x-2} = 9$
III) $\lim_\limits{x \to -\infty}\dfrac{P(x)}{(x+7)^6} = 0$
Find explicit formula for $p(x)$?
A) Calculation:
1) By the rule that if the degree of the numerator is less than the degree of the denominator $\lim_\limits{x \to -\infty}=0$ holds for III. Therefore, $P(x)$ is less than a 6th degree polynomial.
2) Factoring the denominator of II leads to $(x+2)(x-1)$. Therefore, there is a discontinuity at $x=-2$ and $x=1$ for $P(x)$.
B) Further more:
3) I tried to make a polynomial in the numerator which would divide out the discontinuities in the denominator so made a polynomial of $(x+2)(x-1)^3$ which seems to work for II and III on graphing calculator.
4) In addition, the limits of I and II are factors of $3$. This leads me to believe there is a factor in the numerator which will involve $3$.
It should be $$P(x)=(x-1)^3(x+2)(ax+b),$$ which gives the system: $$-2a+b=-\frac{1}{3}$$ and $$a+b=2.$$ I got $$P(x)=\frac{1}{9}(x-1)^3(x+2)(7x+11).$$