#### Answer

(a) $f(0) = \frac{5}{4}$
(b) $\lim\limits_{x \to \infty}f(x) = 5$

#### Work Step by Step

$(x-4)$ is a term in the denominator since $x=4$ is a vertical asymptote.
$(x-1)$ is a term in the numerator since $x=1$ is an x-intercept.
$(x+1)$ is a term in both the numerator and the denominator since there is a removable discontinuity at $x = -1$
We can write the function $f$:
$f(x) = \frac{c~(x-1)~(x+1)}{(x-4)(x+1)}$
where $c$ is a constant
$\lim\limits_{x \to -1}f(x) = \frac{c~[(-1)-1]}{(-1)-4} = \frac{2c}{5} = 2$
Thus, $c = 5$
We can write the function $f$:
$f(x) = \frac{5~(x-1)~(x+1)}{(x-4)(x+1)}$
(a) We can find $f(0)$:
$f(0) = \frac{5~(0-1)~(0+1)}{(0-4)(0+1)} = \frac{5}{4}$
(b) We can find $\lim\limits_{x \to \infty}f(x)$:
$\lim\limits_{x \to \infty}f(x) = \lim\limits_{x \to \infty}\frac{5~(x-1)~(x+1)}{(x-4)(x+1)} = \lim\limits_{x \to \infty}\frac{5~(x^2-1)}{x^2-3x-4} = 5$