Evaluate $\lim_{x \rightarrow 0} x^x$

Solution 1:

You are right, you need to take a logarithm. In this case, $$x^x=e^{x\ln x}$$ so the limit can be written as (by continuity of the exponential function) - writing $\exp x$ for $e^x$, $$\exp{\lim_{x\to 0^+}{x\ln x}}=\exp{\lim_{x\to 0^+}{\frac{\ln x}{1/x}}}=\exp \lim_{x\to 0^+}{\frac{1/x}{-1/x^2}}=\exp\lim_{x\to 0^+}{-x}=e^0=1$$ So the limit is $1$. Amazing!

Solution 2:

The limit is $1$. To see why, let $y = x^x$. then $\ln y = x \ln x = \frac{ \ln x }{\frac{1}{x}} $. Hence

$$ \lim_{x \to 0^+ } \frac{ \ln x }{\frac{1}{x}} =_{L'hop} \lim_{x \to 0^+} \frac{\frac{1}{x} }{ - \frac{1}{x^2}} = \lim_{x \to 0^+ } - x = 0 . $$

Hence,

$$ \lim_{x \to 0^+ } y = e^0 = 1$$

Solution 3:

If you accept that $0^0=1$ , then maybe another approach: since the map $z=f(x,y)=y^x$ is continuous for $y>0$, the limit as $(x,y)\rightarrow (0,0)=0^0=1$