Explanation for a exercise using the L'Hôpital's rule So, I have this equation:
$$
\lim_{x\to0^+} x^x
$$
and the solution:
$$
\lim_{x\to0^+} (e^{\log x})^x = \lim_{x\to0^+} e^{x\log x} = e^{\lim_{x\to0^+}x\log x} = 1
$$
What I don't understand is why is $$\lim_{x\to0^+}x^x = \lim_{x\to0^+}e^{x\log x}$$
Thank you a lot!
 A: $$a^x=b, a\neq 1, a>0\to x=\log_a(b)$$ This means that under proper conditions the functions $f(x)=a^x,~~ g(x)=\log_a(x)$ are inverse to each other. Moreover,  there is a simple rule for logarithm that says $a\log(b)=\log(b^a), b>0$. If you are asking about $$\lim\exp(*)=\exp(\lim(*))$$ I can say that we are allowed to do this cause the function $\exp(x)$ is always continuous over real line.
A: Because $x > 0$ in the limit, $\log (x) $ is well-defined. From the rule $$a^b = e^{b \log a}$$ we get (with $a=b=x$) $$x^x = e^{x \log x}$$
A: See from the logarithmic laws we can write a= e^(ln(a)) [where ln is nothing but the log to the base 'e'.when the base is 'e', the log can be written as 'ln' it is called natural logarithm or the nepoleonic logarithm it is implied that the base is 'e' we just write ln3, means log of 3 to the base 'e'
Now if we write e^(ln 3), it is nothing but 3.
e is powerd by the number (ln3) right! it is nothing but 3 is the answer.
 let us prove that: let
                              X=e^(ln 3)
                     or, ln(X)=ln(e^(ln 3))          [taking ln on both side]

                     or, ln(X)=(ln 3)*(ln(e))             [note that lne is equal to 1]

                     or, ln(X)= (ln 3)*1

                     or, X=3                     [removing ln from both side]

we assumed X= e^(ln 3) and here we got X=3.
so, 
     e^(ln 3) is nothing but actually 3! :-)
**similarly you can understand x^x can be written as e^(ln (x^x)) or e^(x*ln x)) as we know that ln (3^2) is nothing but 2*(ln 3). 
I think this discussion will help you.best of luck.
