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) $\displaystyle\lim_{x\to\infty} h \left(x\right)=\infty$ and
(iii) $\displaystyle\lim_{x\to-1}h(x)=\frac{1}{2}$
The value of $h (1)$ is ______

Answer 1

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