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

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.
