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

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

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

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

- \infty

Solution steps

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

Apply the Limit Chain Rule:

- \infty

= - \infty

x \to 2 lim (\frac{x - 2}{x ^{2} + 4})

x \to - 2 lim (x^{5} + 8 x - 3 x + \frac{7}{9})

x \to π lim (+ (\frac{x}{sin ( x )}))

x \to 1 + lim (x^{l n (l n (x))})

x \to 0 + lim (\frac{2}{sin ( x )})

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