How to calculate $\lim\limits_{x\to 0}\ x^x$?

Question

How do we calculate:

$$\lim_{x \to 0} x^x$$

Answer 1

Rewrite $x^x=e^{x\ln x}$. Now you can approach the limit in the exponent as you would any other limit. Even l'Hospital works just fine if you write it as $\ln x/(1/x)$.

Answer 2

By definition of exponentiation and basic properties of logarithms, $$x^x = \exp(\log(x^x)) = \exp(x\log x ).$$ From continuity of $\exp$ we have $$\lim_{x\to0^+}\exp(x\log x ) = \exp\left(\lim_{x\to 0^+}x\log x\right). $$ By a nice argument from @user17662 here, we have that $$\lim_{x\to0^+}x\log x=0. $$ Hence $$\lim_{x\to0^+}x^x=\exp(0)=1.$$

Answer 3

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

Next, compute the limit of $x\ln(x)$ using L'hopital's.

Answer 4

A different approach to this question is as follows.

First, let's consider $\lim_{n\to\infty}n^{1/n}$. Since $(1+x)^n\ge\dfrac{n(n-1)}{2}x^2>\dfrac{n^2}{4}x^2$, for $n>2$, let $1+x=n^{1/n}$, we have $n\ge\dfrac{n^2}{4}(n^{1/n}-1)^2$. Hence, $0<n^{1/n}-1<\dfrac{2}{\sqrt{n}}$ . We conclude $\lim_{n\to\infty}n^{1/n}=1$.

Therefore, $\lim_{x\to0+}x^x=\lim_{y\to+\infty}\dfrac{1}{y^{1/y}}$ can be estimate by the following inequality:$$[y]^{1/([y]+1)}<y^{1/y}<([y]+1)^{1/[y]}.$$

Show that $\lim_{y\to+\infty}y^{1/y}=1$.

Answer 5

Since the function $f(x)=x^x$ is uniformly continuous in a right neighbourhood of zero, it is enough to replace $x$ with $\frac{1}{n}$ and compute the limit: $$ \lim_{n\to +\infty}\left(\frac{1}{n}\right)^{\frac{1}{n}}=\lim_{n\to +\infty}\frac{1}{n^{1/n}}=1.$$ For instance, the previous line follows from AM-GM: $$ 1\leq n^{1/n} = \left(\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\right)^{1/n} \leq\frac{n+H_{n-1}}{n}\leq 1+\frac{\gamma+\log n}{n}.$$

Answer 6

Hint:

Write $x^x=e^{xln(x)}$ and use the continuity of $e^x$ and l'hospitals rule.

Answer 7

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

Now, applying L'Hospital's Rule,

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