Minimum of x^x on (0, 1]

Worked Solution

How to Think About It: The function $x^x$ looks exotic, but the standard move is to rewrite it as $e^{x \ln x}$, which converts it into a composition you can differentiate cleanly. Before doing any calculus, think about boundary behavior: at $x = 1$ we get $f(1) = 1$, and as $x$ gets small (say $x = 0.1$), we get $0.1^{0.1} \approx e^{0.1 \cdot (-2.3)} \approx e^{-0.23} \approx 0.79$. So the function dips below 1 somewhere in the interior. The limit as $x \to 0^+$ is a classic -- it hinges on the fact that $x \ln x \to 0$.

Quick Estimate: At $x = 1/e \approx 0.368$, we get $f(1/e) = (1/e)^{1/e} = e^{-1/e} \approx e^{-0.368} \approx 0.692$. That is lower than $f(0.1) \approx 0.79$ and $f(1) = 1$, so $x = 1/e$ looks like the minimum. For the limit, $x \ln x$ at $x = 0.01$ gives $0.01 \cdot (-4.6) = -0.046$, which is close to zero, so $f(x) \approx e^{-0.046} \approx 0.955$, heading toward 1.

Approach: Rewrite as an exponential and differentiate.

Formal Solution:

Write $f(x) = e^{x \ln x}$. Differentiating:

$$f'(x) = e^{x \ln x}(\ln x + 1) = x^x(\ln x + 1)$$

Since $x^x > 0$ for all $x > 0$, the sign of $f'(x)$ depends entirely on $\ln x + 1$:

• $f'(x) = 0$ when $\ln x = -1$, i.e., $x = 1/e$
• $f'(x) < 0$ for $x < 1/e$ (function is decreasing)
• $f'(x) > 0$ for $x > 1/e$ (function is increasing)

So $x = 1/e$ is a global minimum on $(0, 1]$:

$$f(1/e) = (1/e)^{1/e} = e^{-1/e} \approx 0.6922$$

For the limit as $x \to 0^+$, we need $\lim_{x \to 0^+} x \ln x$. This is a $0 \cdot (-\infty)$ form. Rewrite as:

$$\lim_{x \to 0^+} \frac{\ln x}{1/x}$$

This is $-\infty / \infty$, so apply L'Hopital's rule:

$$\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$$

Therefore $\lim_{x \to 0^+} x^x = e^0 = 1$.

Answer: The minimum value is $e^{-1/e} \approx 0.6922$, attained at $x = 1/e$. The limit as $x \to 0^+$ is $1$.