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

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

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limit as x approaches 0 of (8^x-2^x)/x

x \to 0 lim (\frac{8 ^{x} - 2 ^{x}}{x})

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

x \to 0 lim (\frac{8 ^{x} - 2 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 2 ln (2)

x \to \infty lim (\frac{arctan ( 4 x )}{8})

x \to 0 lim (\frac{x}{sec ( 4 x )})

h \to 0 lim (\frac{h}{e ^{h} - 1})

x \to - 3 lim (\frac{x ^{2} - 2}{x + 1})

x \to 0 - lim ((1 - e^{x})^{2 x})

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