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

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

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

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

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

Decimal

0.38918 \dots

Solution steps

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

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

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

Apply L'Hopital's Rule

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

Plug in the value

x = 0

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

Simplify

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

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

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