Limit of $\frac{x^c-c^x}{x^x-c^c}$ as $x \rightarrow c$ 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?
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,


$\dfrac{(x^c-c^c)}{x-c} \;$ tends to the derivative of  $x^c$ at $x=c$, i.e. to $c^c$;
$\dfrac{(c^x-c^c)}{x-c}\; $ tends to the derivative of $c^x$ , i.e. $c^c\ln c$.
$\dfrac{(x^x-c^c)}{x-c}\; $ tends to the derivative of $x^x$, i.e.$\;c^c(\ln c+1)$.


Thus $\;\lim\limits_{x\to c}\dfrac{x^c-c^x}{x^x-c^c}=\dfrac{c^c(1-\ln c)}{c^c(1+\ln c)}=\dfrac{1-\ln c}{1+\ln c}$.
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.