Question

Given $f\left(x\right)=\frac{ax+b}{x+1},\underset{x\to \infty }{lim}f\left(x\right)=1$ and $\underset{x\to 0}{lim}f\left(x\right)=2$, then $f(-2)$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Given, $\underset{x\to \infty }{lim}f\left(x\right)=1$

$\Rightarrow \underset{x\to \infty }{lim}\frac{ax+b}{x+1}=1$

$\Rightarrow \underset{x\to \infty }{lim}\frac{a+\frac{b}{x}}{1+\frac{1}{x}}=1$
$\Rightarrow a=1$

Also, $\underset{x\to 0}{lim}f\left(x\right)=2$

$\Rightarrow \underset{x\to 0}{lim}\frac{ax+b}{x+1}=2$

$\Rightarrow b=2$

Now, $f(-2)=\frac{a(-2)+b}{(-2)+1}=\frac{-2+2}{-2+1}=0$