My question is: Show that $\lim_{x \rightarrow c} \frac{x^c-c^x}{x^x-c^c}$ exists and find its value.
Because the limit is 0/0 I've tried using L'Hopital's rule, but every time I differentiate it I still get 0/0?
We're given this hint in the question: use the fact that $a^b=\exp(b\log(a))$ and I've tried doing this to be able to differentiate and use L'Hopital's rule, and also for trying to rearrange it. Can anyone help please?
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Why turn to L'Hopital at all, whose hypotheses are not so easy to check anyway? Rather consider the "simple" derivatives of the functions $f(x)=x^c$, $g(x)=c^x$ and $h(x)=x^x$, namely, $$f'(x)=cx^{c-1}\qquad g'(x)=c^x\ln c\qquad h'(x)=(1+\ln x)x^x$$ Using these at $x=c$, one gets the limits $$\lim_{x\to c}\frac{x^c-c^c}{x-c}=f'(c)=c^c$$ $$\lim_{x\to c}\frac{c^x-c^c}{x-c}=g'(c)=c^c\ln c$$ $$\lim_{x\to c}\frac{x^x-c^c}{x-c}=h'(c)=(1+\ln c)c^c$$ Now, for every $x\ne c$,$$\frac{x^c-c^x}{x^x-c^c}=\frac{\frac{x^c-c^c}{x-c}-\frac{c^x-c^c}{x-c}}{\frac{x^x-c^c}{x-c}}$$ thus, the desired limit is $$\lim_{x\to c}\frac{x^c-c^x}{x^x-c^c}=\frac{f'(c)-g'(c)}{h'(c)}=\frac{1-\ln c}{1+\ln c}$$
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You can apply L'hopital rule, there's nothing wrong in it. Why reinvent the wheel? Lets differentiate wrt $x$: $$\lim_{x\to c}\frac{cx^{c-1}-c^x \ln(c)}{xx^{x-1}+x^x \ln(x)} = \frac{1-\ln(c)}{1+\ln(c)}$$
Here we have used that $\frac{d(x^x)}{dx} = x\cdot x^{x-1} + x^x \ln(x)$, which can be directly evaluated (i dont remember the name of the theorem) and also by taking logarithms.
No need invoke L'Hospital here, just rewrite the fraction: $$\frac{x^c-c^x}{x^x-c^c}=\frac{(x^c-c^c)-(c^x-c^c)}{x-c}\frac{x-c} {(x^x-c^c)}$$ By definition,
Thus $\;\lim\limits_{x\to c}\dfrac{x^c-c^x}{x^x-c^c}=\dfrac{c^c(1-\ln c)}{c^c(1+\ln c)}=\dfrac{1-\ln c}{1+\ln c}$.