I know how to solve this problem by using L'Hospital's rule
$$\lim \limits_{ x\rightarrow 0 }{ { x }^{ x } } =\lim\limits _{ x\rightarrow 0 }{ { e }^{ x\ln { x } } } =\lim\limits _{ x\rightarrow 0 }{ { e }^{ \frac { \ln { x } }{ \frac { 1 }{ x } } } } =\lim\limits _{ x\rightarrow 0 }{ { e }^{ \frac { \frac { 1 }{ x } }{ \frac { 1 }{ { x }^{ 2 } } } } } ={ e }^{ 0 }=1,$$
what other different ways can you suggest or show.thanks
If we set $x=e^{-t}$, then $x\to0^+$ as $t\to\infty$ and $$ x^x=e^{\large-te^{-t}}\tag{1} $$ Since $1+t\le e^t$ for all $t\in\mathbb{R}$, we get, for $t\ge0$, $$ \frac{t^2}4\lt\left(1+\frac t2\right)^2\le e^t\tag{2} $$ Thus, for $t\gt0$, $$ 0\lt te^{-t}\lt\frac4t\tag{3} $$ Thus, $$ e^{\large-\frac4t}\le e^{-te^{-t}}\le1\tag{4} $$ Let $t\to\infty$, and the Squeeze Theorem says that $$ \lim_{x\to0^+}x^x=\lim_{t\to\infty}e^{\large-te^{-t}}=1\tag{5} $$
It is a basic fact, seen in high school, that $\lim_{x\to0_+}x\ln x=0$. Hence $x^x$ tends to $1$ as $x\to 0_+$
As pointed by Bernard, that is the same as showing that $$ \lim_{x\to 0^+} x\log x = 0 \tag{1}$$ or, by setting $x=e^{-t}$, $$ \lim_{t\to +\infty} t e^{-t} = 0\tag{2} $$ that follows by squeezing: for any $t>0$, $e^t>1+t+\frac{t^2}{2}$, hence: $$ 0\leq \lim_{t\to +\infty} te^{-t} \leq \lim_{t\to +\infty}\frac{t}{1+t+\frac{t^2}{2}}=0.\tag{3}$$
Another option is to use the inequality $$\log x < x - 1\tag{1}$$ for $x > 1$. If $x > 1$ then $\sqrt{x} > 1$ and hence $$0 < \log \sqrt{x} < \sqrt{x} - 1 < \sqrt{x}$$ or $$0 < \log x < 2\sqrt{x}$$ or $$0 < \frac{\log x}{x} < \frac{2}{\sqrt{x}}$$ Taking limits as $x \to \infty$ and using Squeeze theorem we get $$\lim_{x \to \infty}\frac{\log x}{x} = 0\tag{2}$$ Letting $t = 1/x$ we get $$\lim_{t \to 0^{+}}t\log t = 0\tag{3}$$ which is equivalent to $$\lim_{t \to 0^{+}}t^{t} = 1$$ as desired.
