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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.

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    $\begingroup$ I'd carry out $y$ first, letting $x=ty,\,t\to 1$ $\endgroup$ – Alexey Burdin May 24 '20 at 20:38
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$$\frac{x^y-y^x}{x^x-y^y}=\frac{\dfrac{x^x-y^x}{x-y}-\dfrac{x^x-x^y}{x-y}}{\dfrac{x^x-y^y}{x-y}}$$ 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, $$\lim_{x\to y}\frac{x^y-y^x}{x^x-y^y}=\lim_{x\to y}\frac{xy^{x-1}-x^y\log x}{y^y(1+\log y)}=\frac{y^y(1-\log y)}{y^y(1+\log y)}=\frac{1-\log y}{1+\log y}\,.$$

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  • $\begingroup$ It may be worth editing in a demonstration that, as $x\to y$, this $\to\frac{1-\ln y}{1+\ln y}$. $\endgroup$ – J.G. May 24 '20 at 20:53
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    $\begingroup$ Got it thanks... Seeing the limit $x\to y$ I actually did think that it might be useful to use the definition of differentiation but couldn't visualize exactly what was coming so didn't continue.. Moral: Never give up! Thank you once again. $\endgroup$ – Dhrubajyoti Bhattacharjee May 24 '20 at 21:06
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    $\begingroup$ @DhrubajyotiBhattacharjee Glad I could help :) $\endgroup$ – A. Goodier May 24 '20 at 21:09
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    $\begingroup$ One thing that I think I should point out is that: Since $$\lim_{x\to y}\frac{f(x) -f(y)}{x-y}=f'(y)$$ is the derivative of $f(x)$ at $x=y.$ So $$\lim_{x\to y}\frac{\frac{x^x-y^x}{x-y}-\frac{x^x-x^y}{x-y}}{\frac{x^x-y^y}{x-y}}=\frac{{f_1}'(y) -{f_2}'(y)}{{f_3}'(y)}$$ i.e., we no longer have $$\lim_{x\to y}$$ and it's a bit unusual that here as $y$ is treated as a constant so we would consider a function of $y$ rather than $x$ such as $f_1(y)=y^x.$ I have observed that if we start with $\frac{x^y-y^x}{x-y}=\frac{\frac{x^y-y^y}{x-y}-\frac{y^x-y^y}{x-y}}{\frac{x^x-y^y}{x-y}},$ [to be continued ] $\endgroup$ – Dhrubajyoti Bhattacharjee May 25 '20 at 5:30
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    $\begingroup$ (continue) then we don't face such issues. Because in this case, we get: $$\lim_{x\to y}\frac{x^y-y^x}{x-y}=\frac{{f_1}'(y) -{f_2}'(y)}{{f_3}'(y)}, $$ where $f_1(x) =x^y, f_2(x) =y^x$ and $f_3(x) =x^x.$ So now, ${f_1}'(x) =yx^{y-1}\implies {f_1}'(y) =y\cdot y^{y-1}=y^y$ and so on. So here it's pretty obvious that the limit $=\frac{1-\ln y}{1+\ln y}.$ Your concept was great but please consider these things.. Thank you! $\endgroup$ – Dhrubajyoti Bhattacharjee May 25 '20 at 5:42
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Write $x^y=e^{(y\ln x)}$, then factor the numerator and denominator.

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