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)

Popular >

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 Notation

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

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

= \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)

x \to \frac{π}{4} lim (\frac{4 x}{tan ( x )})

x \to 2 lim (\frac{7}{x - 1})

x \to 3 - lim ((x) + (4 - x))

x \to 1 lim (ln (x) ln (x - 1))

x \to 0 lim (\frac{5 x}{3 sin ( x ) cos ( x )})

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)