If \(\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}\) exists and equal to 2…
If \(\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}\) exists and equal to 2 , then the ordered pair \((\alpha, \beta)\) of real numbers is
\((1,-1)\)
\((-2,1)\)
\((-1,1)\)
\((1,-2)\)
Solution
It is given that,
\(\begin{aligned}
& \lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}=2 \\
& \Rightarrow \lim _{x \rightarrow \infty} \frac{x^3+1-\alpha x^3-\beta x^2-\alpha x-\beta}{x^2+1}=2
\end{aligned}\)
For the existance of limit, coefficient of \(x^3=0\)
\(\begin{aligned}
& \therefore \alpha=1 \\
& \therefore \lim _{x \rightarrow \infty} \frac{-\beta x^2-x-\beta+1}{x^2+1}=2 \\
& \Rightarrow \lim _{x \rightarrow \infty} \frac{-\beta-\frac{1}{x}-\frac{\beta}{x^2}+\frac{1}{x^2}}{1+\frac{1}{x^2}}=2 \Rightarrow-\beta=2 \Rightarrow \beta=-2 \\
& \therefore(\alpha, \beta)=(1,-2)
\end{aligned}\)
Hence, option (4) is correct.