Let $0 \leq l < s < u $. I want to show 2 things:
1) The function $x \mapsto \frac{u^x-s^x}{u^x-l^x}$ defined for $x \in \mathbb{R}\setminus{0}$ is increasing.
2) $\lim_{x\to0} \frac{u^x-s^x}{u^x-l^x} = \frac{\ln(u)-\ln(s)}{\ln(u)-\ln(l)}$.
Both these things can be verified on https://www.desmos.com/calculator/trhaqobquo.
$1)$ Mainly in such situations we count first derivative and here we do it, too (although intuitively we see that this function must be increasing)
$$f'(x)=\left(\dfrac{u^x-s^x}{u^x-l^x}\right)'=\dfrac{(u^x\ln{u}-s^x\ln{s})(u^x-l^x)-(u^x-s^x)(u^x\ln{u}-l^x\ln{l})}{(u^x-l^x)^2}=\dfrac{(u^x)^2 \ln{u}-l^xu^x\ln{u}-s^xu^x\ln{s}+s^xl^x\ln{s}-\left((u^x)^2\ln{u}-u^xl^x\ln{l}-s^xu^x\ln{u}+s^xl^x\ln{l}\right)}{(u^x-l^x)^2}=\dfrac{l^x u^x(\ln{l}-\ln{u})+s^x u^x(\ln{u}-\ln{s})+s^xl^x(\ln{s}-\ln{l})}{(u^x-l^x)^2}=\dfrac{(ul)^x\ln{\frac{l}{u}}+(su)^x\ln{\frac{u}{s}}+(sl)^x\ln{\frac{s}{l}}}{(u^x-l^x)^2}.$$ We can observe that always $f'(x)>0$, because of the assumptions.
$2)$ L'Hospital's rule will be helpful, beacuse this is limit of a type $\left[\frac{0}{0}\right]$
$$\lim_{x\to 0}\dfrac{u^x-s^x}{u^x-l^x}\stackrel{H}{=}\lim_{x\to 0}\dfrac{u^x\ln{u}-s^x\ln{s}}{u^x\ln{u}-l^x\ln{l}}=\dfrac{\ln{u}-\ln{s}}{\ln{u}-\ln{l}}$$