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

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)

x \to - 1 lim (2 x - 5)

t \to 4 + lim (t^{2} + 5 t - 35)

x \to 2 lim ((6 x + 3)^{2})

x \to 3 lim ((5 - 4 x)^{2})

z \to 9 lim (\frac{9 z - z ^{2}}{z - 3})

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)