What is the limit as x approaches 0 of (7^x-4^x)/x ?

The limit as x approaches 0 of (7^x-4^x)/x is ln(7)-2ln(2)

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limit as x approaches 0 of (7^x-4^x)/x

x \to 0 lim (\frac{7 ^{x} - 4 ^{x}}{x})

ln (7) - 2 ln (2)

Decimal Notation

0.55961 \dots

Solution steps

x \to 0 lim (\frac{7 ^{x} - 4 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= ln (7) - 2 ln (2)

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What is the limit as x approaches 0 of (7^x-4^x)/x ?
The limit as x approaches 0 of (7^x-4^x)/x is ln(7)-2ln(2)