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

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

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

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

1

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= 1

\frac{1}{( s + 1 ) ^{\frac{1}{2}}}

(- 6, 4), (- 1, 4)

\frac{d}{d x} (\frac{x ^{n - 1}}{n})

\int_{x}^{2 x} y d y

n = 0 \sum \infty \frac{n ^{3}}{5 ^{n}}

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