$\lim _\limits{x \rightarrow 0} \dfrac{a^x-b^x}{c^x-d^x}=$

The given limit is $L = \lim_{x \rightarrow 0} \dfrac{a^x - b^x}{c^x - d^x}$.

Since the limit is of the form $\dfrac{0}{0}$ as $x \rightarrow 0$, we apply L'Hopital's rule by differentiating the numerator and the denominator with respect to $x$.

The derivative of $a^x$ is $a^x \ln a$.

Applying the rule:

$L = \lim_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}(a^x - b^x)}{\dfrac{d}{dx}(c^x - d^x)} = \lim_{x \rightarrow 0} \dfrac{a^x \ln a - b^x \ln b}{c^x \ln c - d^x \ln d}$.

Evaluating the limit as $x \rightarrow 0$:

$L = \dfrac{a^0 \ln a - b^0 \ln b}{c^0 \ln c - d^0 \ln d} = \dfrac{\ln a - \ln b}{\ln c - \ln d}$.

Using the property of logarithms $\ln x - \ln y = \ln(\dfrac{x}{y})$:

$L = \dfrac{\ln(\dfrac{a}{b})}{\ln(\dfrac{c}{d})}$.

This is equivalent to $\dfrac{\log(\dfrac{a}{b})}{\log(\dfrac{c}{d})}$.

Answer: $\dfrac{\log \left(\dfrac{a}{b}\right)}{\log \left(\dfrac{c}{d}\right)}$