What is the value of $\lim_{x\to 0}x^x$? 
Evaluate $$\lim_{x\to 0}x^x$$

I tried by writing $x$ in terms of exponentials:
$x=e^{\ln x}$ so $x^x=e^{x\ln x}$
$\lim_{x \to 0}x \ln x=\lim_{x \to 0}(\ln x +1) =-\infty$
Thus $\lim_{x\to 0}x^x=0$
Is this correct?
 A: It is correct to first calculate $\lim_{x \to 0}x\ln (x)=L$ and then, since $\exp$ is continuous the desire result will be $e^L$, but why do you think $x \ln (x) = \ln(x)+1$? That it is not true. Better use that 
$$
x \ln (x) = \frac{\ln(x)}{1/x}
$$
Now apply L´Hopital's rule to get that $ \lim_{x \to 0}x \ln (x)= -0 $, and conclude that $\lim_{x \to 0}x^x =1$ .
A: Apply L'hopitale rule: 
$x\ln x = \dfrac{\ln x}{\dfrac{1}{x}}\to \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^2}}=-x \to 0^{-}\Rightarrow x^x \to e^{0}=1$
A: Here is a way that does not involve L'Hopital's rule. Consider the logarithm $f(x)=x\ln x$. Write $x=\mathrm e^{-y}$, so that $x\to0$ means $y\to\infty$. Then $$f(x)=-\frac{y}{\mathrm e^{y}}.$$Just from the third term $\frac12y^2$ of the exponential series, we can see that this tends to $0$ (on the negative side) as $y\to\infty$. So the limit of the original expression $x^x=\exp f(x)$ is $1$.
A: Put $y = x^x$. Then $\log(y) = x\log(x)$. As $x\downarrow 0$, we have a $0\cdot \infty$ situation.  So we rearrange to get
$$\log(y) = {\log(x)\over 1/x}$$  As $x\to 0$ we have $\infty/\infty$ so we can invoke L'hospital to get
$$\lim_{x\downarrow 0} y = {1/x\over  -1/x^2} = \lim_{x\downarrow 0} -x = 0.$$  
Therefore the limit is 1.  Caveat: You must take the limit as $x\downarrow 0$, or you have domain problems.
