What is the limit as x approaches 1 of ln(x^2-1) ?

The limit as x approaches 1 of ln(x^2-1) is Does not exist

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limit as x approaches 1 of ln(x^2-1)

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

Does not exist

Solution steps

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

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

then the limit does not exist

x \to 1 + lim (ln (x^{2} - 1)) = - \infty

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

Does not exist

= Does not exist

\frac{1}{s ^{2} + 3 s}

x \to π - lim ((x + π) csc (x))

\frac{d}{d x} (2 x e^{x} - e^{2 x})

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

f (x) = \frac{x}{1 + x ^{2}}, (3, 0.3)

What is the limit as x approaches 1 of ln(x^2-1) ?
The limit as x approaches 1 of ln(x^2-1) is Does not exist