What is the limit as n approaches 0 of (((3)^n-1))/n ?

The limit as n approaches 0 of (((3)^n-1))/n is ln(3)

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limit as n approaches 0 of (((3)^n-1))/n

n \to 0 lim (\frac{( ( 3 ) ^{n} - 1 )}{n})

ln (3)

Decimal

1.09861 \dots

Solution steps

n \to 0 lim (\frac{3 ^{n} - 1}{n})

Apply L'Hopital's Rule

: n \to 0 lim (\frac{ln ( 3 ) \cdot 3 ^{n}}{1})

= n \to 0 lim (\frac{ln ( 3 ) \cdot 3 ^{n}}{1})

Plug in the value

n = 0

= \frac{ln ( 3 ) \cdot 3 ^{0}}{1}

Simplify

= ln (3)

x \to 0 + lim (ln (1 - cos (x)) - ln (sin (x)))

\frac{d}{d x} (18 x)

\frac{d y}{d t} + 2 y = 1, y (0) = \frac{5}{2}

\frac{\partial}{\partial z} (x y - 3 x z + yz)

\frac{d ^{3}}{d x ^{3}} (x^{4} - 3 x^{2} + 6 x - 5)

What is the limit as n approaches 0 of (((3)^n-1))/n ?
The limit as n approaches 0 of (((3)^n-1))/n is ln(3)