What is the limit as x approaches 0 of (7^x-9^x)/x ?

The limit as x approaches 0 of (7^x-9^x)/x is ln(7)-2ln(3)

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limit as x approaches 0 of (7^x-9^x)/x

x \to 0 lim (\frac{7 ^{x} - 9 ^{x}}{x})

ln (7) - 2 ln (3)

Decimal Notation

- 0.25131 \dots

Solution steps

x \to 0 lim (\frac{7 ^{x} - 9 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{7 ^{x} ln ( 7 ) - 2 ln ( 3 ) \cdot 9 ^{x}}{1})

= x \to 0 lim (\frac{7 ^{x} ln ( 7 ) - 2 ln ( 3 ) \cdot 9 ^{x}}{1})

Plug in the value

x = 0

= \frac{7 ^{0} ln ( 7 ) - 2 ln ( 3 ) \cdot 9 ^{0}}{1}

Simplify

\frac{7 ^{0} ln ( 7 ) - 2 ln ( 3 ) \cdot 9 ^{0}}{1} : ln (7) - 2 ln (3)

= ln (7) - 2 ln (3)

g (t) = 4 t^{- \frac{1}{8}}

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\int_{(- 2)}^{8} (∣ x - 2 ∣ - 3) d x

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\int_{0}^{\frac{13}{6}} π (\frac{13}{6} - x)^{2} d x

What is the limit as x approaches 0 of (7^x-9^x)/x ?
The limit as x approaches 0 of (7^x-9^x)/x is ln(7)-2ln(3)