What is the limit as x approaches-1 of ln(1+x) ?

The limit as x approaches-1 of ln(1+x) is Does not exist

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limit as x approaches-1 of ln(1+x)

x \to - 1 lim (ln (1 + x))

Does not exist

Solution steps

x \to - 1 lim (ln (1 + x))

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

then the limit does not exist

x \to - 1 + lim (ln (1 + x)) = - \infty

x \to - 1 - lim (ln (1 + x)) =

Does not exist

= Does not exist

\frac{d}{d x} (ln (4 x - 5))

\frac{1}{1 - x - x ^{2}}

t - 2

f (x) = x^{2} + 4, at x = - 1

\int \frac{( x ^{2} - x + 18 )}{x ^{3} + 3 x} d x

What is the limit as x approaches-1 of ln(1+x) ?
The limit as x approaches-1 of ln(1+x) is Does not exist