# How do you find the limit of x^x as x>0^-?

##### 1 Answer
Dec 14, 2017

${\lim}_{x \to {0}^{+}} {x}^{x} = 1$

#### Explanation:

Write the function as:

${x}^{x} = {\left({e}^{\ln} x\right)}^{x} = {e}^{x \ln x}$

Note now that:

${\lim}_{x \to {0}^{+}} x \ln x$

is in the indeterminate form $0 \cdot \infty$. We can reconduce it to the form $\frac{\infty}{\infty}$ and then apply l'Hospital's rule:

${\lim}_{x \to {0}^{+}} x \ln x = {\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{x}} = {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} \frac{1}{x}} = {\lim}_{x \to {0}^{+}} \frac{\frac{1}{x}}{- \frac{1}{x} ^ 2} = {\lim}_{x \to {0}^{+}} - x = 0$

As ${e}^{x}$ is a continuous function near $0$:

${\lim}_{x \to {0}^{+}} {e}^{x \ln x} = {e}^{{\lim}_{x \to {0}^{+}} x \ln x} = {e}^{0} = 1$

graph{x^x [-10, 10, -5, 5]}