Hey I have to study the existence of the limit of this function at $(0,0)$:
$f:(0,+\infty)\times(0,+\infty)\to\Bbb R $ $$f(x,y)= \begin{cases} \frac{e^x -e^y}{\log(x)-\log(y) } & \ x \neq y\\ xe^x & x=y \end{cases}$$
I tried to find curves where the limits were different, but I couldn't find any: all that I tried tend to zero. So I tried to delimit the function, but I couldn't do it either.
I think the limit does not exist but I don't know how to prove it.
We have to precise that it is more a question of continuity in $(0,0)$
We have to show in the will of showing f is continuous that the limit goes to $ f(0,0)=0 $
Use Mean Value Theorem :
Let $x\neq y$,
$$ h:x\to e^x$$ $$g:x\to \ln(x)$$
$$ \exists c_1\in]x,y[ \ , h'(c_1) =\dfrac{\ln(x)-\ln(y)}{x-y} $$ $$ \exists c_2\in]x,y[ \ , g'(c_2) =\dfrac{e^x-e^y}{x-y} $$
So you have , because of $c_1,c_2\to 0$ when $(x,y) \to 0. $ (Use $ |c_{1,2}-x|<|x-y|$)
$$ f(x,y)=\frac{g'(c_2)}{h'(c_1)}=\frac{e^x-e^y}{\ln(x)-\ln(y)}=\dfrac{c_2}{e^{c_1}}_{(x,y)\to(0,0)}\to0$$
Then: $$ \lim_{(x,y)\to(0,0)} f(x,y) = \lim_{(c_1,_2)\to(0,0)} \frac{c_2}{e^{c_1}} = 0 $$ Thus continuity.