As $x\to 0$ from the right, $\ln \left(x\right)\to -\infty$, so the product $x \ln \left(x\right)$ tends to the indeterminate form $0\cdot \infty$. This form is converted to either $0\/0$ or $\infty \/\infty$ by algebraically rewriting the product as a quotient, then applying L'Hôpital's rule. The full solution is as follows.

$\underset{x\to 0\+}{lim}x \ln \left(x\right)\=\underset{x\to 0\+}{lim}\frac{\ln \left(x\right)}{\frac{1}{x}}\=\underset{x\to 0\+}{lim}\frac{\frac{1}{x}}{-\frac{1}{x^2}}\=\underset{x\to 0\+}{lim}\left(-x\right)\=0$ 

If the product is written as the quotient $\frac{x}{\frac{1}{\ln \left(x\right)}}$, L'Hôpital's rule would apply, but it would be very difficult to obtain the limit of the resulting expression.

Alternatively, see Tables 3.9.7(a - d).$$

Click  tutor to launch the Limit Methods tutor for this example. See Figure 3.9.7(a) for the relevant initial state of the tutor. Note the selection of "right" as the Direction.

Click the L'Hôpital's rule button to rewrite $x \ln \left(x\right)$ as $\frac{\ln \left(x\right)}{1\/x}$. As per Figure 3.9.7(b), select $x$ under the "Numerator:" heading, and click the right-facing arrow. The result is seen in Figure 3.9.7(c) where $\ln \left(x\right)$ is listed as the numerator, and $1\/x$ is listed as the denominator. The display shows the limit in a form for which L'Hôpital's rule applies.$$

Click the "Apply" button to obtain the configuration in Figure 3.9.7(d).

All that remains is to apply the Constant Multiple and Identity rules to obtain $\underset{x\to 0\+}{lim}x \ln \left(x\right)\=0$.