What is the limit as t approaches 0 of ((3^t-2^t))/t ?

The limit as t approaches 0 of ((3^t-2^t))/t is ln(3)-ln(2)

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limit as t approaches 0 of ((3^t-2^t))/t

t \to 0 lim (\frac{( 3 ^{t} - 2 ^{t} )}{t})

ln (3) - ln (2)

Decimal

0.40546 \dots

Solution steps

t \to 0 lim (\frac{3 ^{t} - 2 ^{t}}{t})

Apply L'Hopital's Rule

: t \to 0 lim (\frac{3 ^{t} ln ( 3 ) - 2 ^{t} ln ( 2 )}{1})

= t \to 0 lim (\frac{3 ^{t} ln ( 3 ) - 2 ^{t} ln ( 2 )}{1})

Plug in the value

t = 0

= \frac{3 ^{0} ln ( 3 ) - 2 ^{0} ln ( 2 )}{1}

Simplify

\frac{3 ^{0} ln ( 3 ) - 2 ^{0} ln ( 2 )}{1} : ln (3) - ln (2)

= ln (3) - ln (2)

\int \frac{( 3 x - 2 )}{1 - 6 x - 9 x ^{2}} d x

\frac{\partial}{\partial y} (\frac{( - y ^{2} )}{( x - y ) ^{2}})

x \to 5 - lim (\frac{2}{x})

x \to 2 lim (\frac{x ^{2} + 6 x - 2 x}{x - 2})

\frac{1}{x ^{5}} \frac{d y}{d x} = 2 + 18 y

What is the limit as t approaches 0 of ((3^t-2^t))/t ?
The limit as t approaches 0 of ((3^t-2^t))/t is ln(3)-ln(2)