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 π lim (\frac{x}{6} - \frac{π}{2})

x \to 2 - lim (e^{\frac{3}{x - 2}})

x \to 0 lim (\frac{sin ( 2 \cdot x )}{x})

t \to 0 + lim ((t - 2)^{5})

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)