What is the limit as x approaches 0 of ((7^x-8^x))/x ?

The limit as x approaches 0 of ((7^x-8^x))/x is ln(7)-3ln(2)

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limit as x approaches 0 of ((7^x-8^x))/x

x \to 0 lim (\frac{( 7 ^{x} - 8 ^{x} )}{x})

ln (7) - 3 ln (2)

Decimal Notation

- 0.13353 \dots

Solution steps

x \to 0 lim (\frac{7 ^{x} - 8 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{7 ^{x} ln ( 7 ) - 3 ln ( 2 ) \cdot 8 ^{x}}{1})

= x \to 0 lim (\frac{7 ^{x} ln ( 7 ) - 3 ln ( 2 ) \cdot 8 ^{x}}{1})

Plug in the value

x = 0

= \frac{7 ^{0} ln ( 7 ) - 3 ln ( 2 ) \cdot 8 ^{0}}{1}

Simplify

\frac{7 ^{0} ln ( 7 ) - 3 ln ( 2 ) \cdot 8 ^{0}}{1} : ln (7) - 3 ln (2)

= ln (7) - 3 ln (2)

x \to 0 lim (x^{2} e \frac{1}{x ^{2}})

x \to \infty lim (\frac{( 5 ^{x} + x ^{4} )}{( 7 ^{x} + x ^{2} )})

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

x \to π lim (sin^{2} (x))

x \to 0 lim (x^{2} + 3 x - 4)

What is the limit as x approaches 0 of ((7^x-8^x))/x ?
The limit as x approaches 0 of ((7^x-8^x))/x is ln(7)-3ln(2)