Limit of function formed by taking difference after swapping a variable / constant How to apply L'Hospital's Rule to evaluate limit at $x=a$ for 
$$ \frac {f(x,a)-f(a,x)}{(x-a)}  \tag{1} $$
which has $\frac00$ form? If it cannot be applied, how is the limit found?
EDIT1:

The blue and purple plots are for curves:
$$ y=\dfrac{ x^e - e^x}{ x - e}; \,  y=  (x^e - e^x)  $$
respectively for the limit and the numerator, which are both vanishing at $x=e.$
EDIT2
If $ f(x,a) = a \sin x$ or $x \sin a, $ then such a limit is $ \dfrac{\partial f}{\partial x}(a,a)-\dfrac{\partial f}{\partial a}(a,a)= a \cos a - \sin a, $
If $ f(x,a) = a ^x $ or $x^a$ ,then such a limit is $ \dfrac{\partial f}{\partial x}(a,a)-\dfrac{\partial f}{\partial a}(a,a)= a^a (\log a-1). $
EDIT3:
So we can define for two variables an order independent Swap derivative for (1) as
$$ SDf(x,a) = \dfrac{\partial f}{\partial x}(a,a)-\dfrac{\partial f}{\partial a}(a,a). $$
We have for two example functions  $$ f(x,a) = a \sin x,\;  x^a  $$
their corresponding SD's: 
$$a \cos a - \sin a, \, a^a (\log a-1). $$
 A: Hint: add $f(a,a)$ and subtract $f(a,a)$ in the nominator. Then see the partial derivatives that form. You get $\partial_1f(a,a)-\partial_2 f(a,a)$
$$\lim\limits_{x\to a}\frac{f(x,a)-f(a,a)-(f(a,x)-f(a,a))}{x-a}=\lim\limits_{x\to a}\frac{f(x,a)-f(a,a)}{x-a}-\lim\limits_{x\to a}\frac{f(a,x)-f(a,a)}{x-a}=\frac{\partial f}{\partial x_1}(a,a)-\frac{\partial f}{\partial x_2}(a,a) $$
Example: 
If $f(x,e)=x^e$ then $\frac{\partial f}{\partial x}=ex^{e-1}$ and $\frac{\partial f}{\partial e}=x^e \ln(x)$. Therefore
$$\lim\limits_{x\to e}{\frac{x^e-e^x}{x-e}} =\frac{\partial f}{\partial x}(e,e)-\frac{\partial f}{\partial e}(e,e)=e\cdot e^{e-1}-e^e\ln e=0$$
Remark: As you have the particular type of the function $f(x,a)$ you can already apply L'Hospital's rule also:
$$\lim\limits_{x\to e}{\frac{x^e-e^x}{x-e}} =\lim\limits_{x\to e}{\frac{(x^e-e^x)'}{(x-e)'}}=\lim\limits_{x\to e}{\frac{ex^{e-1}-e^x}{1}}=0$$
L'Hospital's rule, can be applied as follows to the original problem: if you denote $\varphi(x)=f(x,a)$ and $\psi(x)=f(a,x)$ you have in the nominator $(\varphi(x)-\psi(x))'_{|x=a}=\varphi'(x)_{|x=a}-\psi'(x)_{|x=a}=\frac{\partial f}{\partial x_1}(a,a)-\frac{\partial f}{\partial x_2}(a,a)$  
