calculus - L'hopital's Rule on an indeterminate difference

I have the following question, in which I am told to use L'Hopital's rule:

$\lim_{x\to \infty}(-x+5\cdot ln(x))$

From eyeballing it, I would conclude that the polynomial $x$ will decrease faster than the logarithmic $ln$ would increase, meaning the limit would be $-\infty$, but I can't see how to use L'hopitals rule here as I've been told. I know that the idea is to put this into a quotient, but if I do the following:

$$-x+5\cdot ln(x) = \frac{-x^2}{x}+\frac{5x\cdot ln(x)}{x} = \frac{-x^2 + 5x\cdot ln(x)}{x}$$

And then try to take the limit, I get a doubly indeterminate form of $\frac{-\infty+\infty}{\infty}$, and using L'Hopitals rule here just ends back inthe same kind of indeterminate difference.

Answer

$$\begin{array}{lll} \displaystyle\lim_{x\to \infty}(-x+5\ln x)&=&\displaystyle\lim_{x\to \infty}(-x+\ln x^5)\\ &=&\displaystyle\lim_{x\to \infty}(\ln(e^{-x+\ln x^5}))\\ &=&\ln\displaystyle\lim_{x\to \infty}(e^{-x+\ln x^5})\\ &=&\ln\displaystyle\lim_{x\to \infty}(e^{-x}e^{\ln x^5})\\ &=&\ln\displaystyle\lim_{x\to \infty}(x^5e^{-x})\\ &=&\ln\displaystyle\lim_{x\to \infty}\bigg(\frac{x^5}{e^x}\bigg)\\ &=&\dots\\ \end{array}$$