Answer
On the graph, we can see that $~~\frac{5^x-4^x}{3^x-2^x} \approx 0.55~~$ as $~~x \to 0$
$\lim\limits_{x \to 0}\frac{5^x-4^x}{3^x-2^x} = \frac{ln~5-ln~4}{ln~3-ln~2} \approx 0.55034$
Work Step by Step
On the graph, we can see that $~~\frac{5^x-4^x}{3^x-2^x} \approx 0.55~~$ as $~~x \to 0$
$\lim\limits_{x \to 0}\frac{5^x-4^x}{3^x-2^x} = \lim\limits_{x \to 0}\frac{1-1}{1-1} = \frac{0}{0}$
We can apply L'Hospital's Rule.
$\lim\limits_{x \to 0}\frac{ (ln~5)~5^x-(ln~4)~4^x }{ (ln~3)~3^x-(ln~2)~2^x} = \frac{ (ln~5)~(1)-(ln~4)~(1) }{ (ln~3)~(1)-(ln~2)~(1)} = \frac{1.609438-1.386294}{1.098612-0.693147} = 0.55034$
Therefore:
$\lim\limits_{x \to 0}\frac{5^x-4^x}{3^x-2^x} = \frac{ln~5-ln~4}{ln~3-ln~2} \approx 0.55034$