Answer
$f_n(x)\rightarrow \sqrt x$
Work Step by Step
Consider the functions:
$f_{n+1}(x)=\dfrac{1}{2}\left(f_n(x)+\dfrac{x}{f_n(x)}\right)$
Compute $f_n(x)$ for $n=3,4,5$:
$f_3(x)=\dfrac{1}{2}\left(\dfrac{1}{2}(x+1)+\dfrac{x}{\dfrac{1}{2}(x+1)}\right)=\dfrac{x^2+6x+1}{4(x+1)}$
$f_4(x)=\dfrac{1}{2}\left(\dfrac{x^2+6x+1}{4(x+1)}+\dfrac{x}{\dfrac{x^2+6x+1}{4(x+1)}}\right)=\dfrac{x^4+28x^3+70x^2+28x+1}{8(x+1)(x^2+6x+1)}$
$f_5(x)=\dfrac{1}{2}\left(\dfrac{x^4+28x^3+70x^2+28x+1}{8(x+1)(x^2+6x+1)}+\dfrac{x}{\dfrac{x^4+28x^3+70x^2+28x+1}{8(x+1)(x^2+6x+1)}}\right)$
$=\dfrac{x^8+120x^7+1820x^6+8008x^5+12,870x^4+8008x^3+1820x^2+120x+1}{16(x+1)(x^2+6x+1)(x^4+28x^3+70x^2+28x+1)}$
Graph $f_1(x),f_2(x),f_3(x),f_4(x),f_5(x)$ and $\sqrt x$.
Notice that as $n$ grows, $f_n$ gets closer and closer to $\sqrt x$, therefore $f_n(x)$ is asymptotic to $\sqrt x$.
