Evaluate: $\lim_{x\to y} \frac{x^y-y^x}{x^x-y^y},$ without using L'Hospital's rule. I found this question in my younger brother's mathematics text book and they are not allowed to use L'Hospitals rule in exams. They evaluate limits using some fundamental formulas such as 

$$\lim_{x\to0}\frac{\sin x}x=1,\;\lim_{x\to 0}\frac{e^x-1}{x}=1,\;\lim_{x\to0}\frac{a^x-1}{x}=\ln a,\; \lim_{x\to0}(1+x)^{\frac 1x}=e\text{ or }\lim_{x\to\infty}\left(1+\frac 1x\right)^x=e.$$

Though he knows about L'Hospital's rule but for exam purposes he asked me to solve it without using the same. But I'm not able to solve it using those formulas. So I posted it here to know if there's a way to solve it without the L'Hospital's rule. Thank you in advance.
$$\frac{x^y-y^x}{x^x-y^y}=\frac{\dfrac{x^x-y^x}{x-y}-\dfrac{x^x-x^y}{x-y}}{\dfrac{x^x-y^y}{x-y}}$$
and by definition
$$f'(y)=\lim_{x\to y}\frac{f(x)-f(y)}{x-y}$$
so
$$\lim_{x\to y}\frac{\dfrac{x^x-y^x}{x-y}-\dfrac{x^x-x^y}{x-y}}{\dfrac{x^x-y^y}{x-y}}=\lim_{x\to y}\frac{f_1'(y)-f_2'(y)}{f_3'(y)}$$
where $f_1(y)=y^x, f_2(y)=x^y, f_3(y)=y^y$.

Now, $f_1'(y)=xy^{x-1}, f_2'(y)=x^y\log x$ and $f_3'(y)=y^y(1+\log y)$. 

Therefore,
$$\lim_{x\to y}\frac{x^y-y^x}{x^x-y^y}=\lim_{x\to y}\frac{xy^{x-1}-x^y\log x}{y^y(1+\log y)}=\frac{y^y(1-\log y)}{y^y(1+\log y)}=\frac{1-\log y}{1+\log y}\,.$$
Write $x^y=e^{(y\ln x)}$, then factor the numerator and denominator.