Evaluate $\underset{x\to 0+}{\mathop{\lim }}\,{{\left( {{x}^{x}} \right)}^{x}}$ We’ve already proven $\underset{x\to 0+}{\mathop{\lim }}\,{{x}^{x}}=1$ in the classroom.
Here’s my quick and dirty attempt: $\underset{x\to 0+}{\mathop{\lim }}\,{{\left( {{x}^{x}} \right)}^{x}}={{\left( \underset{x\to 0+}{\mathop{\lim }}\,{{x}^{x}} \right)}^{0}}={{1}^{0}}=1$. Have I pulled this off?
What would be the simplest way to correctly evaluate this limit? I can think of $\underset{x\to 0+}{\mathop{\lim }}\,{{\left( {{x}^{x}} \right)}^{x}}={{e}^{\underset{x\to 0+}{\mathop{\lim }}\,x\log {{x}^{x}}}}$, $\underset{x\to 0+}{\mathop{\lim }}\,x\log {{x}^{x}}=\left( \underset{x\to 0+}{\mathop{\lim }}\,x \right)\left( \underset{x\to 0+}{\mathop{\lim }}\,\log {{x}^{x}} \right)=\left( \underset{x\to 0+}{\mathop{\lim }}\,x \right)\left( \log \underset{x\to 0+}{\mathop{\lim }}\,{{x}^{x}} \right)=0\cdot \log 1=0$, so $\underset{x\to 0+}{\mathop{\lim }}\,{{\left( {{x}^{x}} \right)}^{x}}={{e}^{0}}=1$. This is my second attempt.
 A: $$\lim_{x\to 0^{+}}(x^{x})^{x} = \lim_{x\to 0^{+}}x^{x^{2}}=\lim_{x\to 0^{+}}(x^{2})^{\frac{x^{2}}{2}}\ \stackrel{u=\frac{x^{2}}{2}}{=} \lim_{u\to 0^{+}}(2u)^{u}=\lim_{u\to 0^{+}}2^{u}\lim_{u\to 0^{+}}u^{u}=2^{0}\cdot 1 = \boxed{1}$$
A: The first way is fine, no real need for overcomplication. The exponentiation $(a,b)\mapsto a^b$ is continuous on $(0,\infty)\times \Bbb R$, therefore as long as $g(x)\to \alpha\in\Bbb R$ and $f(x)\to\beta>0$, you can evaluate the limit $f(x)^{g(x)}\to \beta^\alpha$ by substitution.
A: I would use L'Hospital's rule. The original limit is of the indeterminate form $0^0$ so we let
$$L=\lim_{x\to 0^+}(x^x)^x=\lim_{x\to 0^+}x^{x^2}.$$
Then
$$\ln L = \lim_{x\to 0^+}\frac{\ln(x)}{\frac{1}{x^2}},$$
which is now of the indeterminate form $-\infty/\infty.$ Applying L'Hospital's rule gives
$$\ln L = \lim_{x\to 0^+}\frac{\frac 1x}{\frac {-2}{x^3}}=\lim_{x\to 0^+}{\frac{x^2}{-2}}=0.$$
Therefore
$$L = e^{\ln L}=e^0 = 1.$$
A: $$\lim_{x\rightarrow 0^+}\left ( x^x \right )^x=\lim_{x\rightarrow 0^+}\operatorname{exp}\left \{ \ln\left ( x^x \right )^x \right \}=\lim_{x\rightarrow 0^+}\operatorname{exp}\left \{ x\ln \left ( x^x \right ) \right \}=\operatorname{exp}\left \{ 0\ln \left ( \operatorname{exp}\left \{ \lim_{x\rightarrow 0^+}\frac{\ln x}{1/x} \right \} \right ) \right \}=\operatorname{exp}\left \{ 0 \right \}=1$$
