Find $\lim_{x\to e}\frac{x^{e^x}-e^{x^e}}{x-e}$ without L'Hospital

Question

By L'Hospital's rule, it is not hard to find that $$\lim_{x\to e}\frac{x^{e^x}-e^{x^e}}{x-e}=e^{e^e+e-1}.$$ But how to find $\displaystyle\lim_{x\to e}\frac{x^{e^x}-e^{x^e}}{x-e}$ without L'Hospital's rule ?

Thank you.

Answer 1

Hint:

$$x^{e^x} - e^{x^e} = (x^{e^x} - e^{e^e}) -(e^{x^e} - e^{e^e})$$

$$\left( \text{ Use: } \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a) \text{ twice } \right)$$