Find lim as x to 0+ x^alphaln x, where alpha > 0

Find $\displaystyle\lim_{x\to0^{+}}x^{\alpha}\ln x$, where $\alpha\gt 0$.

Step 1: Identify the indeterminate form

As $x\to0^{+}$: $x^{\alpha}\to0$ because $\alpha\gt 0$, while $\ln x\to-\infty$. The product is of the form $0\cdot(-\infty)$.

Step 2: Rewrite as a fraction

\[ x^{\alpha}\ln x = \frac{\ln x}{x^{-\alpha}}. \tag{1} \]

As $x\to0^{+}$: $\ln x\to-\infty$ and $x^{-\alpha}\to+\infty$, so (1) is of the form $\dfrac{-\infty}{+\infty}$.

Step 3: Apply l'Hôpital's rule

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

and the quotient of the derivatives collapses to a single power: \[\frac{x^{-1}}{-\alpha\,x^{-\alpha-1}} = \frac{x^{-1+\alpha+1}}{-\alpha} = -\frac{x^{\alpha}}{\alpha}.\]

\[ \lim_{x\to0^{+}}\frac{\ln x}{x^{-\alpha}} = \lim_{x\to0^{+}}\left(-\frac{x^{\alpha}}{\alpha}\right). \]

Step 4: Evaluate

Since $\alpha\gt 0$ we have $x^{\alpha}\to0$, hence

\[ \lim_{x\to0^{+}}\left(-\frac{x^{\alpha}}{\alpha}\right) = -\frac{0}{\alpha} = 0. \]

Step 5: Verification

\[\alpha = \tfrac12:\quad x = 0.01:\ -0.4605170; \qquad x = 10^{-4}:\ -0.0921034; \qquad x = 10^{-8}:\ -0.0018421 \;\;\checkmark\]

\[\text{The minimum is } -\frac{1}{\alpha e} \text{ at } x = e^{-1/\alpha};\ \text{for } \alpha = \tfrac12:\ -0.7357589 \text{ at } 0.1353353 \;\;\checkmark\]

\[\text{Substituting } x = u^{1/\alpha}:\ x^{\alpha}\ln x = \frac{1}{\alpha}\,u\ln u \to 0 \text{ by Problem 391} \;\;\checkmark\]