What is the limit as x approaches 0 of (9^x-2^x)/x ?

The limit as x approaches 0 of (9^x-2^x)/x is 2ln(3)-ln(2)

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limit as x approaches 0 of (9^x-2^x)/x

x \to 0 lim (\frac{9 ^{x} - 2 ^{x}}{x})

2 ln (3) - ln (2)

Decimal Notation

1.50407 \dots

Solution steps

x \to 0 lim (\frac{9 ^{x} - 2 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 ln ( 3 ) \cdot 9 ^{x} - 2 ^{x} ln ( 2 )}{1})

= x \to 0 lim (\frac{2 ln ( 3 ) \cdot 9 ^{x} - 2 ^{x} ln ( 2 )}{1})

Plug in the value

x = 0

= \frac{2 ln ( 3 ) \cdot 9 ^{0} - 2 ^{0} ln ( 2 )}{1}

Simplify

\frac{2 ln ( 3 ) \cdot 9 ^{0} - 2 ^{0} ln ( 2 )}{1} : 2 ln (3) - ln (2)

= 2 ln (3) - ln (2)

\frac{d}{d x} (7 (x^{3}) - x)

\int_{1}^{1.5} cos (\frac{π x}{3}) d x

\frac{\partial}{\partial y} (2 e^{y} + 3 sin (y))

x^{2}, at x = 1

\frac{\partial}{\partial x} (\frac{1}{r ^{2}})

What is the limit as x approaches 0 of (9^x-2^x)/x ?
The limit as x approaches 0 of (9^x-2^x)/x is 2ln(3)-ln(2)