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^{2} - 1

is positive when

x \to 1 +

. Therefore

x^{2} - 1 = x^{2} - 1

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

Apply the Limit Chain Rule:

- \infty

= - \infty

\int_{0}^{1} x 5 x + 4 d x

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

\frac{d}{d x} (x n (x))

\int \frac{1}{v ^{2} + 3 v - 4} d v

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