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 -infinity

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

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

- \infty

Solution steps

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

Apply the Limit Chain Rule:

- \infty

= - \infty

\int x (x - 1) d x

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 6)^{7})

g (x) = sin (x), 0, \frac{π}{4}

\frac{d}{d x} ((x^{2} + 1) (x^{2} - 1))

n = 2 \sum \infty \frac{n}{2 ^{n - 1}}

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 -infinity