[tex]a_{n+1}=\dfrac{a_n+\frac4{a_n}}2[/tex]
Assuming convergence of the sequence, we have [tex]a_n\to L[/tex] for some limit [tex]L[/tex] as [tex]n\to\infty[/tex]. This also means [tex]a_{n+1}\to L[/tex]. So in the above equation, we can substitute and solve for [tex]L[/tex]:
[tex]L=\dfrac{L+\frac4L}2[/tex]
[tex]\implies2L^2=L^2+4[/tex]
[tex]\implies L^2=4\implies L=\pm2[/tex]
The limit can only be one of these values; otherwise, the sequence would be divergent. To decide which one is correct, we can observe that the sequence is strictly positive. So [tex]L=2[/tex].