The sequence has the property that \(\displaystyle a_n= \frac{a_{n-1}}{2}+ \frac{1}{a_{n-1}}\). Assuming the sequence has a limit, call it A. Of course, \(\displaystyle \{a_n\}\) and \(\displaystyle \{a_{n-1}\}\) are just different ways of numbering that same sequence so \(\displaystyle \lim_{n\to\infty} a_n= A\) and \(\displaystyle \lim_{n\to\infty} a_{n-1}= A\). Taking the limit of both sides of \(\displaystyle a_n= \frac{a_{n-1}}{2}+ \frac{1}{a_{n-1}}\), as afos suggested, gives \(\displaystyle A= \frac{A}{2}+ \frac{1}{A}\) (as long as \(\displaystyle A\ne 0\)). Solve that for A. Once you have a value for A, try showing that this is a decreasing sequence having A (or a number less than A) as a lower bound.