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

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

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

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

2 ln (2) - ln (7)

Decimal Notation

- 0.55961 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 2 ln (2) - ln (7)

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