What is the limit as x approaches 0 of (e^x-1)/(8x) ?

The limit as x approaches 0 of (e^x-1)/(8x) is 1/8

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limit as x approaches 0 of (e^x-1)/(8x)

x \to 0 lim (\frac{e ^{x} - 1}{8 x})

\frac{1}{8}

Decimal

0.125

Solution steps

x \to 0 lim (\frac{e ^{x} - 1}{8 x})

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

= \frac{1}{8} \cdot x \to 0 lim (\frac{e ^{x} - 1}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{e ^{x}}{1})

= \frac{1}{8} \cdot x \to 0 lim (\frac{e ^{x}}{1})

Plug in the value

x = 0

= \frac{1}{8} \cdot \frac{e ^{0}}{1}

Simplify

\frac{1}{8} \cdot \frac{e ^{0}}{1} : \frac{1}{8}

= \frac{1}{8}

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What is the limit as x approaches 0 of (e^x-1)/(8x) ?
The limit as x approaches 0 of (e^x-1)/(8x) is 1/8