If $P\left(x\right)$ is a polynomial such that

$P\left(x\right)+P\left(2x\right)=5x^2-18$ then $\underset{x\to 3}{lim} \frac{P\left(x\right)}{x-3}$ is

Solution:

Clearly, $P\left(x\right)$ is a quadratic polynomial. So, let

$P\left(x\right)=a{x}^2+bx+c$ Then

$\begin{array}{l}P\left(x\right)+P\left(2x\right)=5x^2-18\\ \Rightarrow a{x}^2+bx+c+4a{x}^2+2bx+c=5x^2-18\\ \Rightarrow 5a{x}^2+3bx+2c=5x^2-18\\ \Rightarrow 5a=5,3b=0\text{ and }2c=-18\\ \Rightarrow a=1,b=0\text{ and }c=-9\\ \therefore P\left(x\right)=x^2-9\end{array}$

and , Hence $\underset{x\to 3}{lim} \frac{P\left(x\right)}{x-3}=\underset{x\to 3}{lim} \frac{x^2-9}{x-3}=\underset{x\to 3}{lim} (x+3)=6$