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

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

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

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

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 2 ln (2)

\frac{d}{d x} (sin^{2} (\frac{x + 2}{x + 1}))

y^{^{'}} = ln (x) y + x^{x}

\frac{d}{d x} (sin (x) + cos (x))

\int_{0}^{2} x^{2} (x^{3} + 1)^{\frac{3}{2}} d x

3 x^{2}, x^{2} + 4, - 2, 2

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