Reverse Coefficient

Note that $g(X) = X^3f(X^{-1})$ so that, for $u \neq 0$ , $f(u) = 0$ if and only if $g(u^{-1}) = 0$ . Since $g(3)=0$ , $f(\tfrac13) = 0$ . Since $\tfrac{f(X)}{g(X)}$ is well-behaved as $X \to 3$ , we also have $f(3) = 0$ . Thus $f(X) \; = \; (X - 4)(X-3)(X-\tfrac13) \hspace{2cm} g(X) = -\tfrac14(X - \tfrac14)(X - \tfrac13)(X - 3)$ and hence $\begin{aligned} f(X) + g(X) & =\; \tfrac34(X+1)(X - \tfrac13)(X - 3) \\ \frac{f(X) + g(X)}{X + 1} & = \; \tfrac34(X - \tfrac13)(X - 3) \end{aligned}$ so that $\lim_{X \to -1} \frac{f(X)+g(X)}{X+1} \; = \; \tfrac34 \times -\tfrac43 \times -4 \; = \; \boxed{4}$

$f(3)=27+9a+3b+c=0$

$g(4)=64a+16b+4a+1=0$

$\displaystyle f(3)=27+9a+3b+c=27\cdot\left(\frac{1}{27}c+\frac{1}{9}b+\frac{1}{3}a+1\right)=27\cdot g\left(\frac{1}{3}\right)=0$

$\displaystyle\therefore g\left(\frac{1}{3}\right)=0$

Since the value of $\displaystyle\lim_{x\rightarrow\frac{1}{3}}\frac{f(x)}{g(x)}$ exists, $\displaystyle f\left(\frac{1}{3}\right)=0$

$\displaystyle f\left(\frac{1}{3}\right)=\frac{1}{27}+\frac{1}{9}a+\frac{1}{3}b+c=0$

Now, $\displaystyle\lim_{x\rightarrow -1}\frac{f(x)+g(x)}{x+1}=\lim_{x\rightarrow -1}\frac{(c+1)x^3+(a+b)x^2+(a+b)x+(c+1)}{x+1}=\lim_{x\rightarrow -1}\frac{(c+1)(x+1)(x^2-x+1)+(a+b)x(x+1)}{x+1}=\lim_{x\rightarrow -1}\left\{(c+1)(x^2-x+1)+(a+b)x\right\}=3c+3-a-b$

If you do the calculation, you will get the answer $4$ .