Question

Behavior at the origin Using calculus and accurate sketches, explain how the graphs of f(x) = xᵖ ln x differ as x → 0⁺ for p = 1/2, 1, and 2.

Accepted Answer

First, recognize that the function is given by $$f(x) = x$$^{p}$$ \ln x$$, and we are interested in the behavior as $$x 	o 0$$^{+}$$. Since $$\ln x$$ tends to $$-\infty$$ as x$$ 	o $$0^{+}$$, the product's behavior depends on the power $$p of$$ $$x. $$Rewrite the function to analyze the limit: $$f(x) = x$$^{p}$$ \ln x = \frac{\ln x}{x$$^{-p}$$}. As$$ $$x 	o 0$$^{+}$$, x$$^{-p}$$ 	o +\infty$$, so this is an indeterminate form of type $$\frac{-\infty}{+\infty}$$, which suggests using L'Hôpital's Rule. Apply L'Hôpital's Rule by setting $$g(x) = \ln x$$ and $$h(x) = x$$^{-p}$$. Compute the derivatives: g$$'(x) = \frac{1}{x}$$ and h$$'(x) = -p $$x^{-p-1}. $$Then, the limit becomes $$\lim_{x 	o 0$$^{+}$$} \frac{g'(x)}{h'(x)} = \lim_{x 	o 0$$^{+}$$} \frac{\frac{1}{x}}{-p x$$^{-p-1}$$}. $$Simplify the expression: $$\frac{\frac{1}{x}}{-p x$$^{-p-1}$$} = \frac{1}{x} \cdot \frac{1}{-p} \cdot x$$^{p+1}$$ = -\frac{1}{p} x$$^{p}$$. As x$$ 	o $$0^{+}$$, $$x^{p}$$ 	o 0$$, so the limit is 0$. This shows that $$\lim_{x 	o $$0^{+}$$} $$x^{p}$$ \ln x = 0$$ for all p$$ \u003e 0$$. To understand how the graphs differ for p$$ = \frac{1}{2}, 1,$$ and 2$, note that the factor x$$^{p}$$ approaches zero at different rates. For smaller p$, x$$^{p}$$ approaches zero more slowly, so the negative logarithmic term dominates more, causing the graph to dip more sharply near zero. For larger p$, x$$^{p}$$ goes to zero faster, so the product approaches zero more gently. Sketching these will show the curve for p$$=\frac{1}{2}$$ dipping down more steeply near zero, while for p=2$ it approaches zero more smoothly.

Behavior at the origin Using calculus and accurate sketches, explain how the graphs of f(x) = xᵖ ln x differ as x → 0⁺ for p = 1/2, 1, and 2.

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Key Concepts

Behavior of Functions Near Zero

Properties of the Natural Logarithm Function

Effect of the Power p in xᵖ