Evaluate $\lim_{x \rightarrow 0} x^x$ I know when evaluated it gives one but I can't figure out how to prove it.  Can anyone help?
I believe it requires L'Hopitals rule and taking a natural log but i cannot figure out the exact math.
 A: 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!
A: 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, 
$$ \lim_{x \to 0^+ } y  = e^0 = 1$$
A: 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 $(x,y)\rightarrow (0,0)=0^0=1$
A: $\displaystyle\lim_{x\to0}x^x~=~\lim_{n\to\infty}\bigg(\frac1n\bigg)^\frac1n~=~\lim_{n\to\infty}\frac1{\sqrt[n]n}~.~$ But how much is $\displaystyle\lim_{n\to\infty}\sqrt[n]n$ ? Let us approach $\infty$ on a logarithmic scale, and see what happens! $~\sqrt[10]{10}=10^{^\tfrac1{10}},\quad\sqrt[100]{100}=10^{^\tfrac2{10^2}},~\sqrt[1000]{1000}=10^{^\tfrac3{10^3}}$, and generally, $\sqrt[10^k]{10^k}=10^{^\tfrac k{10^k}}.~$ But obviously $\displaystyle\lim_{k\to\infty}\frac k{10^k}=0,~$ so our limit is $10^0=1$, and $\dfrac11=1$. $\big($Of course, instead of $10$ we could have used any other number $>1\big)$. I've decided to post this approach mainly because I find it more intuitive.
