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 Does not exist

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

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

Does not exist

Solution steps

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

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

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

Does not exist

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

= Does not exist

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

x \to 2 lim (4 x - 3)

x \to 1 lim ((x + 3) (x^{2} + 1))

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

x \to 2 lim (4 x + 8)

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 Does not exist