What is the limit as x approaches 0 of (2^{2x}-2^{-2x})/(2^{3x)-2^{-3x}} ?

The limit as x approaches 0 of (2^{2x}-2^{-2x})/(2^{3x)-2^{-3x}} is 2/3

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limit as x approaches 0 of (2^{2x}-2^{-2x})/(2^{3x)-2^{-3x}}

x \to 0 lim (\frac{2 ^{2 x} - 2 ^{- 2 x}}{2 ^{3 x} - 2 ^{- 3 x}})

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

x \to 0 lim (\frac{2 ^{2 x} - 2 ^{- 2 x}}{2 ^{3 x} - 2 ^{- 3 x}})

Simplify

\frac{2 ^{2 x} - 2 ^{- 2 x}}{2 ^{3 x} - 2 ^{- 3 x}} : \frac{4 ^{- x} ( 4 ^{x} + 1 )}{8 ^{- x} ( 1 6 ^{x} + 4 ^{x} + 1 )}

= x \to 0 lim (\frac{4 ^{- x} ( 4 ^{x} + 1 )}{8 ^{- x} ( 1 6 ^{x} + 4 ^{x} + 1 )})

Plug in the value

x = 0

= \frac{4 ^{- 0} ( 4 ^{0} + 1 )}{8 ^{- 0} ( 1 6 ^{0} + 4 ^{0} + 1 )}

Simplify

\frac{4 ^{- 0} ( 4 ^{0} + 1 )}{8 ^{- 0} ( 1 6 ^{0} + 4 ^{0} + 1 )} : \frac{2}{3}

= \frac{2}{3}

\frac{d}{d x} (- \frac{8}{x ^{3}} + \frac{2}{x ^{2}})

\int \frac{e ^{a r c t a n (x)} + 1}{1 + x ^{2}} d x

x \to - 1 - lim (\frac{x}{∣ x - 1 ∣})

\int \frac{1}{10 r ^{\frac{1}{2}} - r} d r

\frac{d}{d x} (15 x^{2} + 2 x - 8)

What is the limit as x approaches 0 of (2^{2x}-2^{-2x})/(2^{3x)-2^{-3x}} ?
The limit as x approaches 0 of (2^{2x}-2^{-2x})/(2^{3x)-2^{-3x}} is 2/3