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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Simplify

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

= x \to 0 lim (\frac{e ^{x} - 1 - x}{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{1}{2}

\int x e d x

\frac{d}{d x} ((x + 2) e^{2 x})

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

f (x) = 6 x - 3

\frac{d x}{d y} = \frac{2 y + sec ^{2} ( y )}{2 x}

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