First off, the limit does not exist as neither square root nor log function are defined for negative numbers (recall that a limit must exist from the left and right!). This is why the other posted answer is incorrect and, if anything, should emphasize the importance of not just applying l'Hopital's without checking the conditions first, they are not a minor technicality.
Suppose, on the other hand, that you want the right hand limit instead then we will use the fact that the exponential function grows faster than any polynomial, i.e.,
$\lim_{t\to\infty} \frac{t^n}{e^t} = 0$ for any $n$.
In the same way $e^t$ grows really fast, $1/e^t$ shrinks really quickly. To utilize this substitute $x=e^{-2t}$, then
$$ \lim_{x \to 0^+} \sqrt{x} \log(x)^{2015}
= \lim_{t\to \infty} e^{-t} (-2t)^{2015}
= - 2^{2015} \lim_{t\to\infty} \frac{t^{2015}}{e^t}
= 0 $$
Here we are implicitly using the fact that $e^{-2t}$ is one-to-one and continuous, and that $\sqrt{x}\log(x)^{2015}$ is right continuous at $0$.