What is the limit as x approaches 0-of ln|e^x-1| ?

The limit as x approaches 0-of ln|e^x-1| is -infinity

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limit as x approaches 0-of ln|e^x-1|

x \to 0 - lim (ln ∣ e^{x} - 1 ∣)

- \infty

Solution steps

x \to 0 - lim (ln ∣ e^{x} - 1 ∣)

e^{x} - 1

is negative when

x \to 0 -

. Therefore

∣ e^{x} - 1 ∣ = - e^{x} + 1

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

Apply the Limit Chain Rule:

- \infty

= - \infty

\frac{d}{d x} (3 x + x^{\frac{1}{2}})

y = (x - 4) (x^{2} + 8), (1, - 27)

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

\frac{\partial}{\partial x} (x^{0.2} y^{0.8})

\int (4 x + 3) (2 x^{2} + 3)^{3} d x

What is the limit as x approaches 0-of ln|e^x-1| ?
The limit as x approaches 0-of ln|e^x-1| is -infinity