Different way solving limit $\lim \limits_{ x\rightarrow 0 }{ { x }^{ x } } $ 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

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:
$$ 0\leq \lim_{t\to +\infty} te^{-t} \leq \lim_{t\to +\infty}\frac{t}{1+t+\frac{t^2}{2}}=0.\tag{3}$$
