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

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

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

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

3 ln (2) - ln (7)

Decimal

0.13353 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 3 ln (2) - ln (7)

\frac{d}{d x} (tan^{s e c (\frac{1}{x})} (\frac{1}{x}))

f (x) = x^{4} - 6 x^{3}

\int 7 x^{2} + x x d x

\frac{d y}{d t} = 6 y + cos (3 t), y (0) = \frac{28}{15}

\int 7 5 (8 - 2 x)^{8} d x

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