What is the limit as x approaches 0 of (9^x-4^x)/x ?

The limit as x approaches 0 of (9^x-4^x)/x is 2ln(3)-2ln(2)

Popular >

limit as x approaches 0 of (9^x-4^x)/x

x \to 0 lim (\frac{9 ^{x} - 4 ^{x}}{x})

2 ln (3) - 2 ln (2)

Decimal Notation

0.81093 \dots

Solution steps

x \to 0 lim (\frac{9 ^{x} - 4 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 2 ln (3) - 2 ln (2)

f (x) = ln (x^{9})

\int x y e^{x y} d y

n = 0 \sum \infty \frac{1}{n ^{n}}

\frac{\partial}{\partial x} (e^{- t} cos (π) x)

\int \frac{ln ( \frac{1}{x} )}{x ^{2}} d x

What is the limit as x approaches 0 of (9^x-4^x)/x ?
The limit as x approaches 0 of (9^x-4^x)/x is 2ln(3)-2ln(2)