In other words, to find the limit of \(\displaystyle y= x^{x/2}\), start by finding the limit of \(\displaystyle y= ln(x^{x/2})= \frac{1}{2}x ln(x)\). That is of the form "\(\displaystyle 0\cdot\infty\)" which we can the write as \(\displaystyle \frac{1}{2}\frac{ln(x)}{1/x}\), of the form "\(\displaystyle \infty/\infty\)" to which L'Hopital's rule can be applied.
Because the expontial is continuous, after you have found the limit of ln(y) as, say, L, the limit of of the orginal function, \(\displaystyle x^{x/2}\) is \(\displaystyle e^L\).
