What is the limit as x approaches 0 of ((6^x-4^x))/x ?

The limit as x approaches 0 of ((6^x-4^x))/x is ln(6)-2ln(2)

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limit as x approaches 0 of ((6^x-4^x))/x

x \to 0 lim (\frac{( 6 ^{x} - 4 ^{x} )}{x})

ln (6) - 2 ln (2)

Decimal

0.40546 \dots

Solution steps

x \to 0 lim (\frac{6 ^{x} - 4 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - ln ( 2 ) \cdot 2 ^{2 x + 1}}{1})

= x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - ln ( 2 ) \cdot 2 ^{2 x + 1}}{1})

Plug in the value

x = 0

= \frac{6 ^{0} ln ( 6 ) - ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}{1}

Simplify

\frac{6 ^{0} ln ( 6 ) - ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}{1} : ln (6) - 2 ln (2)

= ln (6) - 2 ln (2)

\frac{9 x}{9 x - 5}

(y + 4) \cdot y^{^{'}} = 2 x y + 4 y

n = 0 \sum \infty (\frac{x}{2})^{n}

(2 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0

\frac{d}{d x} (ln (\frac{cos ( x )}{x}))

What is the limit as x approaches 0 of ((6^x-4^x))/x ?
The limit as x approaches 0 of ((6^x-4^x))/x is ln(6)-2ln(2)