Question

Let $f\left(x\right)=x^3-x^2-3x-1,g\left(x\right)=\left(x+1\right)a$ and $h(x)=\frac{f(x)}{g(x)}$ where h is a rational function such that:
(i) it is continuous everywhere except when $x=-1$,
(ii) $\underset{x\to \infty }{\lim }h\left(x\right)=\infty$ and
(iii) $\underset{x\to -1}{\lim }h(x)=\frac{1}{2}$
The value of $h(1)$ is ______

Answer 1

Answer: -0.5

$h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$
$\Rightarrow h\left(x\right)=\frac{x^3-x^2-3x-1}{\left(x+1\right)a}$....(i)
Now, $\underset{x\to -1}{\lim }h\left(x\right)=\frac{1}{2}$
$\Rightarrow \underset{x\to -1}{\lim }\frac{x^3-x^2-3x-1}{\left(x+1\right)a}=\frac{1}{2}$
$\Rightarrow \frac{2}{a}=\frac{1}{2}$
$\Rightarrow a=4$
$\therefore h\left(x\right)=\frac{x^3-x^2-3x-1}{4\left(x+1\right)}$
$\therefore h\left(1\right)=-1/2$