What is the limit as x approaches 0 of (6^x-8^x)/x ?

The limit as x approaches 0 of (6^x-8^x)/x is ln(6)-3ln(2)

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limit as x approaches 0 of (6^x-8^x)/x

x \to 0 lim (\frac{6 ^{x} - 8 ^{x}}{x})

ln (6) - 3 ln (2)

Decimal Notation

- 0.28768 \dots

Solution steps

x \to 0 lim (\frac{6 ^{x} - 8 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - 3 ln ( 2 ) \cdot 8 ^{x}}{1})

= x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - 3 ln ( 2 ) \cdot 8 ^{x}}{1})

Plug in the value

x = 0

= \frac{6 ^{0} ln ( 6 ) - 3 ln ( 2 ) \cdot 8 ^{0}}{1}

Simplify

\frac{6 ^{0} ln ( 6 ) - 3 ln ( 2 ) \cdot 8 ^{0}}{1} : ln (6) - 3 ln (2)

= ln (6) - 3 ln (2)

y^{^{'}} = \frac{1}{y}

\frac{1}{2 \cdot 5} y + y^{^{'}} = \frac{t}{2}, y (30) = 10

(- 10, 1), (0, - 4)

\int \frac{1}{x ^{\frac{1}{2}} + 1} d x

\frac{\partial}{\partial x} (x y (3 - x - y))

What is the limit as x approaches 0 of (6^x-8^x)/x ?
The limit as x approaches 0 of (6^x-8^x)/x is ln(6)-3ln(2)