Question

If $\underset{x\to \infty }{\lim }f(x)$ exists and is finite and non-zeros and if $\underset{x\to \infty }{\lim }\left(f(x)+\frac{3f(x)-1}{f^2(x)}\right)=3$, then the value of $\underset{x\to \infty }{\lim }f(x)$ is

Answer 1

Answer: 1

$\underset{x\to \infty }{\lim }\left(f(x)+\frac{3f(x)-1}{f^2(x)}\right)=3$
$\Rightarrow \left(\underset{x\to \infty }{\lim }f(x)+\frac{3\underset{x\to \infty }{\lim }f(x)-1}{{\left(\underset{x\to \infty }{\lim }f(x)\right)}^2}\right)=3$
$\Rightarrow \left(y+\frac{3y-1}{y^2}\right)=3$
$\Rightarrow y^3-3y^2+3y-1=0$
$\Rightarrow (y-1{)}^3=0$
$\Rightarrow y=1$