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 )}

\int_{0}^{7} \frac{8 t}{( t - 8 ) ^{2}} d t

x \to 0 - lim (x \cdot sin (\frac{1}{x}))

\frac{d}{d t} (- e^{- t})

\frac{d}{d x} (x^{2} 7^{x})

\int \frac{5 x}{( x ^{2} + 2 ) ^{2}} d x

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))