What is the limit as x approaches 0 of ((8^x-5^x))/x ?

The limit as x approaches 0 of ((8^x-5^x))/x is 3ln(2)-ln(5)

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limit as x approaches 0 of ((8^x-5^x))/x

x \to 0 lim (\frac{( 8 ^{x} - 5 ^{x} )}{x})

3 ln (2) - ln (5)

Solution steps

x \to 0 lim (\frac{8 ^{x} - 5 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{3 ln ( 2 ) \cdot 8 ^{x} - 5 ^{x} ln ( 5 )}{1})

= x \to 0 lim (\frac{3 ln ( 2 ) \cdot 8 ^{x} - 5 ^{x} ln ( 5 )}{1})

Plug in the value

x = 0

= \frac{3 ln ( 2 ) \cdot 8 ^{0} - 5 ^{0} ln ( 5 )}{1}

Simplify

\frac{3 ln ( 2 ) \cdot 8 ^{0} - 5 ^{0} ln ( 5 )}{1} : 3 ln (2) - ln (5)

= 3 ln (2) - ln (5)

\frac{1}{e ^{x^{2}}}

y^{^{'}} = x y^{3} - x y

x^{5} - x^{3} + x, 1

\frac{d y}{d x} = cos (3 x)

\frac{\partial}{\partial x} (2 x^{3} y + 8 y^{3})

What is the limit as x approaches 0 of ((8^x-5^x))/x ?
The limit as x approaches 0 of ((8^x-5^x))/x is 3ln(2)-ln(5)