What is the limit as x approaches 0 of (5^x-1)/(7x) ?

The limit as x approaches 0 of (5^x-1)/(7x) is 1/7 ln(5)

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limit as x approaches 0 of (5^x-1)/(7x)

x \to 0 lim (\frac{5 ^{x} - 1}{7 x})

\frac{1}{7} ln (5)

Decimal Notation

0.22991 \dots

Solution steps

x \to 0 lim (\frac{5 ^{x} - 1}{7 x})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{7} \cdot x \to 0 lim (\frac{5 ^{x} - 1}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{ln ( 5 ) \cdot 5 ^{x}}{1})

= \frac{1}{7} \cdot x \to 0 lim (\frac{ln ( 5 ) \cdot 5 ^{x}}{1})

Plug in the value

x = 0

= \frac{1}{7} \cdot \frac{ln ( 5 ) \cdot 5 ^{0}}{1}

Simplify

\frac{1}{7} \cdot \frac{ln ( 5 ) \cdot 5 ^{0}}{1} : \frac{1}{7} ln (5)

= \frac{1}{7} ln (5)

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What is the limit as x approaches 0 of (5^x-1)/(7x) ?
The limit as x approaches 0 of (5^x-1)/(7x) is 1/7 ln(5)