UnLIMITed power!

$\lim\limits_{x\to 0^{+}}-x\ln(x) = \lim\limits_{x\to 0^{+}}\cfrac{\ln(x)}{\frac{1}{-x}}$

We can apply L'Hopital's rule here $\left(\frac{-\infty}{-\infty}\right)$ and differentiate the top and the bottom to get $\lim\limits_{x\to 0^{+}} \cfrac{\frac{1}{x}}{\frac{1}{x^2}} = \lim\limits_{x\to 0^{+}}x = \boxed{0}$

You can slap everything in the log and get log of the limit as x approaches 0 of 1/x^x. It is a known fact that the limit as x approaches 0 of x^x is 1 so thanks to the log you get 0.

If we use limits - 0*ln(0), 0 times anything is 0

You can use Taylor expansion, if you expend the natural logarithm up to the first order around the point x = lim x-> 0+. You get -5.65 + x you multiply by -x and the result follows

L’hopital’s rule is:

$\lim_{ x\to\ a} f(x) \cdot g(x) = M \cdot N$ , where:

$M = \lim_{x\to\ a} f(x)$

$N = \lim_{x\to\ a} g(x)$

So in the equation given, we can say that $f(x)=-x$ and $g(x)=ln(x)$ .

Now, our $M$ and $N$ are:

$M = \lim_{x\to\ 0^+} -x$

$N = \lim_{x\to\ 0^+} ln(x)$

With the limit being as x approaches zero from the right (as positive numbers approach zero).

Plugging in numbers will tell us that $M=0$ , which means that the product $M \cdot N$ will be zero regardless of the value of $N$ .

So, the answer is zero.