If $g(x)=\frac{P(x)}{x^2-4}$, $\lim_{x\rightarrow\infty}g(x)=5$, and $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$, find P(x) (Sweden 1950) If $g(x)=\frac{P(x)}{x^2-4}$, $\lim_{x\rightarrow\infty}g(x)=5$, and $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$, find P(x)
I have been trying to do this question but I have not succeeded. I have attempted to do it in the following way:
$\lim_{x\rightarrow\infty} \frac{P(x)}{x^2-4}=\lim_{x\rightarrow\infty}\frac{P'(x)}{2x}$ (L'Hospital rule)
$=\lim_{x\rightarrow\infty} \frac{P''(x)}{2}$
Hence we have that $P''(x)=10$.
I also attempted to do something with $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$ however I didn't manage to find anything useful. After a lot of trial and error, I found that an equation which fits the given parameters is: $P(x)=5x^2-11x+2$. Could you please explain to me how to do this question and also explain the intuitive reason behind each step of your reasoning?
 A: L'Hopital's Rule also applies for $x \to 2$. We must have $P(2) = 0$ for $\lim_{x\to2}g(x)$ to exist.
Hence we have $$\lim_{x \to 2} \frac{P'(x)}{2x} = \frac94$$
and thus $P'(2) = 9$.
By $P''(x) = 10$ we have $P'(x) = 10x + c_1$. The above gives $c_1 = -11$.
Now $P(x) = 5x^2-11x+c_2$, and since we must have $P(2) = 0$, we have $c_2 = 2$.
Therefore $P(x)=5x^2-11x+2$.
A: You have $g(x) = \frac{P(x)}{x^2-4}$ and you want to find $P$.
First condition about $\lim_{x \to \infty}$ gives us the rank of $P$ together with coefficient. Indeed, if $P$ is arbitrary polynomial $a_nx^n + ... + a_1x + a_0$ where $a_n \neq 0$ then
$$ \lim_{x \to \infty} \frac{P(x)}{x^2-4} = \begin{cases} 0 & n \in \{0,1\} \\ a_n & n = 2 \\ sign(a_n)\cdot \infty & n \ge 3 \end{cases} $$
Hence we need $n=2$ for limit to be finite, and moreover $a_2=5$.
Hence $P(x) = 5x^2 + bx + c$
The second condition tells us something about $\lim_{x \to 2}$, the point where denominator is $0$, so that (hence limit is finite), numerator needs to be $0$, too. It means that $2$ is a root of $P$, so that $P(x)=5(x-2)(x-d)$
And lastly, the value of that limit as $x\to 2$ gives us:
$$ \frac{9}{4} = \lim_{x \to 2} \frac{5(x-2)(x-d)}{(x-2)(x+2)} = \lim_{x \to 2}\frac{5(x-d)}{(x+2)} = \frac{5(2-d)}{4}$$
From there $2-d = \frac{9}{5}$ and $d = 2-\frac{9}{5} = \frac{1}{5}$
Hence $P(x) = 5(x-2)(x-\frac{1}{5}) = (x-2)(5x-1) = 5x^2 -11x + 2$
A: Since $\lim_{x\to +\infty} g(x)$ and $\lim_{x\to 2} g(x)$ exist, we can clude that P(x) has the following form:
$$g(x)=\frac{(x-2)(ax+b)}{x^2-4}=\frac{ax+b}{x+2}$$
then make use of the condition $\lim_{x\to +\infty} g(x)=5$ and $\lim_{x\to 2} g(x)=\frac94$, we have $a=5$ and $b=-1$.
Finally,
$$P(x)=(x-2)(ax+b)=5x^2-11x+2$$
