What is the limit as x approaches 0 of (e^x-1)/(18x) ?

The limit as x approaches 0 of (e^x-1)/(18x) is 1/18

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limit as x approaches 0 of (e^x-1)/(18x)

x \to 0 lim (\frac{e ^{x} - 1}{18 x})

\frac{1}{18}

Decimal Notation

0.05555 \dots

Solution steps

x \to 0 lim (\frac{e ^{x} - 1}{18 x})

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

= \frac{1}{18} \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}{18} \cdot x \to 0 lim (\frac{e ^{x}}{1})

Plug in the value

x = 0

= \frac{1}{18} \cdot \frac{e ^{0}}{1}

Simplify

\frac{1}{18} \cdot \frac{e ^{0}}{1} : \frac{1}{18}

= \frac{1}{18}

\frac{x}{2 x + x - 3}

\int (in x)^{2} d x

x \to \infty lim (1 + x)

y = 2 (x + 1), y = 3 (x + 1), x = 5

\frac{d}{d x} (x^{2} + y^{3})

What is the limit as x approaches 0 of (e^x-1)/(18x) ?
The limit as x approaches 0 of (e^x-1)/(18x) is 1/18