What is the limit as h approaches 0-of (2^h-1)/h ?

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

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limit as h approaches 0-of (2^h-1)/h

h \to 0 - lim (\frac{2 ^{h} - 1}{h})

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

h \to 0 - lim (\frac{2 ^{h} - 1}{h})

Apply L'Hopital's Rule

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

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

Plug in the value

h = 0

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

Simplify

= ln (2)

x \to 0 lim (- \frac{5}{x})

\int \frac{x + 5}{( x + 3 ) ( x - 1 )} d x

\frac{d}{d x} (15 ln (\frac{x}{80}))

\frac{\partial}{\partial y} (y^{x z} + 2 (yz)^{x})

y = - \frac{4 x}{x ^{2} + 1}

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