Find lim as x to 0+ x ln x

Find $\displaystyle\lim_{x\to0^{+}}x\ln x$.

Step 1: Identify the indeterminate form

As $x\to0^{+}$ we have $x\to0$ and $\ln x\to-\infty$, so the product is of the form $0\cdot(-\infty)$. Nothing can be concluded from the factors separately.

Step 2: Rewrite as a fraction

L'Hôpital's rule applies only to quotients, so invert the factor that tends to zero: $x\ln x = \frac{\ln x}{1/x}$.

As $x\to0^{+}$: $\ln x\to-\infty$ and $1/x\to+\infty$, which is the form $\dfrac{-\infty}{+\infty}$.

Step 3: Apply l'Hôpital's rule

\[ \frac{d}{dx}\bigl[\ln x\bigr] = \frac1x, \qquad \frac{d}{dx}\left[\frac1x\right] = -\frac{1}{x^{2}}, \]

and the new quotient simplifies in one step: $\frac{1/x}{-1/x^{2}} = \frac1x\cdot\bigl(-x^{2}\bigr) = -x$.

Hence $\lim_{x\to0^{+}}\frac{\ln x}{1/x} = \lim_{x\to0^{+}}\bigl(-x\bigr)$.

Step 4: Evaluate

\[ \lim_{x\to0^{+}}\bigl(-x\bigr) = 0. \]

Step 5: Verification

\[x = 0.1:\ -0.2302585; \qquad x = 0.01:\ -0.0460517; \qquad x = 10^{-6}:\ -1.38\cdot10^{-5} \;\;\checkmark\]

\[\text{The product dips to a minimum } -\tfrac1e = -0.3678794 \text{ at } x = \tfrac1e \;\;\checkmark\]

\[\text{Substituting } x = e^{-u}:\ x\ln x = -u\,e^{-u} \to 0 \text{ as } u\to+\infty \;\;\checkmark\]