What is the limit as x approaches 0 of ((3^x-4^x))/x ?

The limit as x approaches 0 of ((3^x-4^x))/x is ln(3)-2ln(2)

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limit as x approaches 0 of ((3^x-4^x))/x

x \to 0 lim (\frac{( 3 ^{x} - 4 ^{x} )}{x})

ln (3) - 2 ln (2)

Decimal Notation

- 0.28768 \dots

Solution steps

x \to 0 lim (\frac{3 ^{x} - 4 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= ln (3) - 2 ln (2)

e^{9 x}

\frac{1}{3} sin (3 x)

12 x^{3} - 3 y^{2} x^{4} + 1 y^{^{'}} = 0, y (0) = 2

\frac{cos ( x )}{x ^{9}}

\int \frac{1}{u ^{2} - u} d u

What is the limit as x approaches 0 of ((3^x-4^x))/x ?
The limit as x approaches 0 of ((3^x-4^x))/x is ln(3)-2ln(2)