Question

$f\left(x\right)=\frac{a{x}^2+b}{x^2+1},\underset{x\to 0}{\lim }f\left(x\right)=1$ and $\underset{x\to \infty }{\lim }f\left(x\right)=1,$ then prove that f(−2) = f(2) = 1

Answer 1

$$\begin{gathered} f\left(x\right)=\frac{a{x}^2+b}{x^2+1} \\ \underset{x\to 0}{\lim }f\left(x\right)=1 \\ \Rightarrow \underset{x\to 0}{\lim }\left[\frac{a{x}^2+b}{x^2+1}\right]=1 \\ \Rightarrow \frac{a\times 0+b}{a+1}=1 \\ \Rightarrow b=1 \\ \mathrm{Also},\underset{x\to \infty }{\lim }f\left(x\right)=1 \\ \underset{x\to \infty }{\lim }\left[\frac{a{x}^2+b}{x^2+1}\right]=1 \end{gathered}$$

Dividing the numerator and the denominator by x²:

$$\begin{gathered} \Rightarrow \underset{x\to \infty }{\lim }\left[\frac{a+{\displaystyle \frac{b}{x^2}}}{1+{\displaystyle \frac{1}{x^2}}}\right]=1 \\ \mathrm{As}x\to \infty ,\frac{1}{x},\frac{1}{x^2}\to 0 \\ \Rightarrow \frac{a+0}{1+0}=1 \\ \Rightarrow a=1 \\ \therefore a=1,b=1 \\ \Rightarrow f\left(x\right)=\frac{x^2+1}{x^2+1}=1 \\ f\left(-2\right)=1\left[\mathrm{Since}f\left(x\right)\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{function},\mathrm{its}\mathrm{value}\mathrm{does}\mathrm{not}\mathrm{depend}\mathrm{on}x.\right] \\ f\left(2\right)=1 \end{gathered}$$