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{\partial}{\partial y} (\frac{x z}{x ^{2} + y ^{2}})

\frac{7}{( s - 1 ) ^{3}}

(10 x + y) d x + (x + 2 y) d y = 0

\int 2 e^{7 x} d x

x \to + 2 lim (\frac{7 x + 4}{3 x + 9})

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