How to find the limit $x=\lim_{a\to{b}}{\frac{a^b-b^a}{a^a-b^b}}$ without using L'Hopital's rule? Let
$$x=\lim_{a\to{b}}{\frac{a^b-b^a}{a^a-b^b}}$$
It is very simple to solve it using L'Hopital's rule, but problem is to solve this limit without L'Hopital's rule. Is there any way to do this?
 A: Let 
$$f(a)=a^b-b^a$$
and 
$$g(a)=a^a-b^b$$
then we have
$$x=\lim_{a\to b}\frac{f(a)-f(b)}{a-b}\times \lim_{a\to b}\left(\frac{g(a)-g(b)}{a-b}\right)^{-1}=\frac{f'(b)}{g'(b)}=\frac{1-\log b}{1+\log b}$$
A: Well, this is essentally the long form of L'Hopital, but:
$$\begin{align}\frac{a^b-b^a}{a^a-b^b} &= \frac{a^b-b^b+b^b-b^a}{a-b} \cdot\frac{a-b}{a^a-b^b}\\
&=\frac{\dfrac{a^b-b^b}{a-b} - \dfrac{b^a-b^b}{a-b}}{\dfrac{a^a-b^b}{a-b}}
\end{align}$$
The three terms here converges to the derivative of the functions $x^b$, $b^x$, and $x^x$ at $x=b$.
A: We can proceed as follows: $$\begin{aligned}L &= \lim_{a \to b}\frac{a^{b} - b^{a}}{a^{a} - b^{b}}\\
&= \lim_{a \to b}\frac{\exp(b\log a) - \exp(a\log b)}{\exp(a\log a) - \exp(b\log b)}\\
&= \lim_{a \to b}\frac{\exp(a\log b)\{\exp(b\log a - a\log b) - 1\}}{\exp(b\log b)\{\exp(a\log a - b\log b) - 1\}}\\
&= \lim_{a \to b}1\cdot\frac{\exp(b\log a - a\log b) - 1}{\exp(a\log a - b\log b) - 1}\\
&= \lim_{a \to b}\frac{\exp(b\log a - a\log b) - 1}{b\log a - a\log b}\cdot\frac{b\log a - a\log b}{a\log a - b\log b}\cdot\frac{a\log a - b\log b}{\exp(a\log a - b\log b) - 1}\\
&= \lim_{a \to b}1\cdot\frac{b\log a - a\log b}{a\log a - b\log b}\cdot 1\\
&= \lim_{h \to 0}\frac{b\log (b + h) - (b + h)\log b}{(b + h)\log (b + h) - b\log b}\\
&= \lim_{h \to 0}\frac{b\log (1 + (h/b)) - h\log b}{b\log (1 + (h/b)) + h\log (b + h)}\\
&= \lim_{h \to 0}\dfrac{\dfrac{\log (1 + (h/b))}{h/b} - \log b}{\dfrac{\log (1 + (h/b))}{h/b} + \log (b + h)}\\
&= \frac{1 - \log b}{1 + \log b}\end{aligned}$$ We have used the fundamental limits $$\lim_{x \to 0}\frac{\log (1 + x)}{x} = 1,\,\lim_{x \to 0}\frac{\exp(x) - 1}{x} = 1$$
A: Let $a={(b+h)}$ so that $h\to0$, then:
$$\eqalign{x=
&=\lim_{a\to{b}}{\frac{a^b-b^a}{a^a-b^b}}\\
&=\lim_{h\to{0}}{\frac{{(b+h)}^b-b^{(b+h)}}{{(b+h)}^{(b+h)}-b^b}}\\
&=\lim_{h\to0}\frac{(1+\frac hb)^b-b^h}{{\left(1+\frac hb\right)}^b.(b+h)^h-1}\\
&\large=\lim_{h\to0}\frac{e^{b\ln(1+\frac hb)}-e^{h\ln b}}{e^{b\ln(1+\frac hb)+h[\ln b+\ln(1+\frac bh)]}-1}=\cdots\\
}$$
Use
$$\lim_{x\to0}\frac{\ln(1+ax)}{ax}=1$$
And,
$$\lim_{x\to0}\frac{e^{ax}-1}{ax}=1$$
Using these two I think you can solve the question as there remained only a calculative part, for now you are conceptually clear.
