Find lim as x to 0+ x^x

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

Step 1: Identify the indeterminate form

As $x\to0^{+}$ the base tends to $0$ and so does the exponent: the form is $0^{0}$. The base alone would suggest the limit $0$, the exponent alone would suggest $1$; neither argument is valid.

Step 2: Rewrite through the exponential

\[ x^{x} = e^{x\ln x}. \tag{1} \]

The exponential function is continuous, so the limit may be moved inside it: \[\lim_{x\to0^{+}}x^{x} = \exp\left(\lim_{x\to0^{+}}x\ln x\right). \tag{2}\]

Everything now depends on the exponent alone.

Step 3: Evaluate the exponent

The product $x\ln x$ is of the form $0\cdot(-\infty)$; written as a quotient it yields to l'Hôpital's rule: \[x\ln x = \frac{\ln x}{1/x}, \qquad \lim_{x\to0^{+}}\frac{\ln x}{1/x} = \lim_{x\to0^{+}}\frac{1/x}{-1/x^{2}} = \lim_{x\to0^{+}}\bigl(-x\bigr) = 0. \tag{3}\]

This is Problem 391.

Step 4: Conclude

Substituting (3) into (2): $\lim_{x\to0^{+}}x^{x} = e^{0} = 1$.

Step 5: Verification

\[x = 0.1:\ 0.7943282; \qquad x = 0.01:\ 0.9549926; \qquad x = 10^{-6}:\ 0.9999862 \;\;\checkmark\]

\[\text{The approach is not monotone: } x^{x} \text{ has a minimum } e^{-1/e} = 0.6922006 \text{ at } x = \tfrac1e \;\;\checkmark\]

\[\bigl(x^{x}\bigr)' = x^{x}\bigl(\ln x+1\bigr) = 0 \iff x = e^{-1} \;\;\checkmark\]