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)

n = 0 \sum \infty 100

y^{^{''}} - 2 y^{^{'}} + y = x^{2}

\frac{\partial}{\partial x} (x^{3} - 5 x y)

\frac{x d y}{d x} + (3 x + 1) y = e^{- 3 x}

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)