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

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}

\int \frac{10 x ^{2} - 9 x + 8}{x ^{3} + 4 x} d x

f (x) = 19 x - 1.86 x^{2}, at x = 1

\frac{\partial}{\partial x} (\frac{10}{x})

f (x) = \frac{2 x}{x + 1}, at x = 1

\frac{d y}{d x} + 4 x y^{2} = 0

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