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

The limit as x approaches 0 of x/((4^x-2^x)) is 1/(ln(2))

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limit as x approaches 0 of x/((4^x-2^x))

x \to 0 lim (\frac{x}{( 4 ^{x} - 2 ^{x} )})

\frac{1}{ln ( 2 )}

Decimal Notation

1.44269 \dots

Solution steps

x \to 0 lim (\frac{x}{4 ^{x} - 2 ^{x}})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= \frac{1}{ln ( 2 )}

\frac{d}{d x} (arcsin (2 x^{5}))

x^{2} + 9

\frac{d}{d x} ((5 x^{3} + 5) (x^{4} - 3 x))

\int [cos (4 x) + cos (8 x)]^{2} d x

\frac{d}{d x} (12 x^{\frac{1}{3}})

What is the limit as x approaches 0 of x/((4^x-2^x)) ?
The limit as x approaches 0 of x/((4^x-2^x)) is 1/(ln(2))