SOLUTION 4: $$ \displaystyle{ \lim_{x \rightarrow 0} \frac{ 3^x-2^x }{ x^2-x } } =
\displaystyle{ \frac{ 3^{0}-2^{0} }{ (0)^2-(0) } } = \displaystyle{ \frac{ 1-1 }{ 0-0 } } = \frac{"0"}{0} $$
(Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately.)
$$ = \displaystyle{ \lim_{x \rightarrow 0} \frac{ 3^x \ln 3 - 2^x \ln 2 }{ 2x-1 } } $$
$$ = \displaystyle{ \frac{ 3^{0} \ln 3 - 2^{0} \ln 2 }{ 2(0)-1 } } $$
$$ = \displaystyle{ \frac{ (1) \ln 3 - (1) \ln 2 }{ -1 } } $$
$$ = \displaystyle{ \ln 2 - \ln 3 } $$
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