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))

Apply the Limit Chain Rule:

- \infty

= - \infty

x \to + 0 lim (\frac{( cos ( m x ) - cos ( n x ) )}{x ^{2}})

x \to 0 lim (4 x - 8)

x \to \frac{1}{2} + lim (x^{2} tan (π x))

x \to - 1 lim (\frac{x}{∣ x + 1 ∣})

x \to - 2 - lim (h (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