$$\displaystyle\lim_{x\to 0^{\hspace{.02 in}+}} \: (x\ln(x)) =\ ?$$
This should be $0$ times $-\infty$, so I believe it's indeterminate in this form, but I don't know how to solve the problem any further than this, if it is possible.
Thank you for any help.
It's zero.
Trivial explanation: the logarithm is a very sloooow function, whereas $x$ is not, hence $x$ dominates.
Non trivial explanation: you can use Hôpital rule after having rewritten it in a smart way as
$$\lim_{x\to 0^+} \frac{\ln(x)}{\frac{1}{x}}$$
A brutal substitution leads you to a Hôpital form $-\infty/\infty$. Applying Hôpital:
$$\lim_{x\to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x\to 0^+} -x = 0$$
You can rewrite this as
$$\lim_{x\to0^+}\frac{\ln x}{\frac1x}$$ and use L'Hospital.
You get
$$\lim_{x\to0^+}\frac{\frac1x}{-\frac1{x^2}}$$
which should be trivial to calculate.
L'Hospital's rule is not the swiss knife of limits computations!
The simplest is to make the $x=\dfrac1t$ ($t\to+\infty$) substitution: $$x\ln x=-\frac{\ln t}t\xrightarrow[t\to+\infty]{}0,$$ the latter being a standard high school limit.