What is the limit as h approaches 0 of (4^h-1)/h ?

The limit as h approaches 0 of (4^h-1)/h is 2ln(2)

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limit as h approaches 0 of (4^h-1)/h

h \to 0 lim (\frac{4 ^{h} - 1}{h})

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

h \to 0 lim (\frac{4 ^{h} - 1}{h})

Apply L'Hopital's Rule

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

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

Plug in the value

h = 0

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

Simplify

= 2 ln (2)

x \to \frac{13 π}{3} lim (cos (x))

\int \frac{e ^{x}}{x} d x

x^{3}, x^{2} + x, - 1, 1.6

\frac{8 x + 1}{x}

f (x) = x^{2} - 2 x + 1, at x = - 1

What is the limit as h approaches 0 of (4^h-1)/h ?
The limit as h approaches 0 of (4^h-1)/h is 2ln(2)