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))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Apply the Limit Chain Rule:

- \infty

= - (- \infty)

Simplify

= \infty

x \to 2 lim (x^{2} - 6 x + 8)

x \to 3 lim (3 π)

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

x \to 0 lim (x - 4)

x \to 5 - lim (\frac{5 x}{x ^{2} - 25})

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