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

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

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

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

\frac{1}{5}

Decimal Notation

0.2

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{1}{5} \cdot \frac{e ^{0}}{1}

Simplify

\frac{1}{5} \cdot \frac{e ^{0}}{1} : \frac{1}{5}

= \frac{1}{5}

y^{^{''}} - 2 y^{^{'}} + y = 3 e^{x}

\int \frac{9 x}{x ^{3} + 1} d x

\frac{\partial}{\partial x} (2 - (x^{2} + y^{2})^{\frac{1}{7}})

y = (x^{5} - 1)^{3}

x \to 0 lim (x sin (\frac{1}{∣ x ∣}))

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