If f (x) = (ax +b) / (x + 1), lim_x→∞ f (x) = 1 and lim_x→0 f (x) = 2, then f (- 2) =

\(\begin{array}{l}\begin{array}{l} \lim _{x \rightarrow \infty} \frac{a x+b}{x+1}=1\\ \text { Divide both numerator \& denominator by } x\\ \lim _{x \rightarrow \infty} \frac{\frac{a x}{x}+\frac{b}{x}}{\frac{x}{x}+\frac{1}{x}}=1\\ \lim _{x \rightarrow \infty} \frac{a+\frac{b}{x}}{1+\frac{1}{x}}=1 \quad \begin{array}{l} x \rightarrow \infty \\ \frac{1}{x} \rightarrow 0 \end{array}\\ \frac{a}{1}=1\\ \therefore a=1\\ \lim _{x \rightarrow 0} f(x)=2\\ \lim _{x \rightarrow 0} \frac{a x+b}{x+1}=2\\ \lim _{x \rightarrow 0} \frac{x+b}{x+1}=2\\ \frac{b}{1}=2\\ b=2\\ f(x)=\frac{x+2}{x+1}\\ f(-2)=\frac{-2+2}{-2+1}=0 \end{array}\end{array} \)