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

The limit as x approaches 0 of 1/(e^x-1)-1/x is -1/2

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

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

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

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

Simplify

\frac{1}{e ^{x} - 1} - \frac{1}{x} : \frac{x - e ^{x} + 1}{x ( e ^{x} - 1 )}

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- e ^{x}}{e ^{x} x + 2 e ^{x}})

= x \to 0 lim (\frac{- e ^{x}}{e ^{x} x + 2 e ^{x}})

Plug in the value

x = 0

= \frac{- e ^{0}}{e ^{0} \cdot 0 + 2 e ^{0}}

Simplify

\frac{- e ^{0}}{e ^{0} \cdot 0 + 2 e ^{0}} : - \frac{1}{2}

= - \frac{1}{2}

\int (\frac{x ^{3} - 2 x}{x}) d x

y = - \frac{1}{2 x ^{3}}, (2, - \frac{1}{16})

x \to 0 lim (2 x ln (1 - x^{2}))

\frac{d}{d x} (\frac{3 x + 6}{2 x + 3})

y^{^{''}} + 80 y^{^{'}} + 1200 y = 0

What is the limit as x approaches 0 of 1/(e^x-1)-1/x ?
The limit as x approaches 0 of 1/(e^x-1)-1/x is -1/2