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

The limit as x approaches-1 of ln|1+x| is -infinity

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limit as x approaches-1 of ln|1+x|

x \to - 1 lim (ln ∣ 1 + x ∣)

- \infty

Solution steps

x \to - 1 lim (ln ∣ 1 + x ∣)

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

then

x \to a lim f (x) = L

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

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

= - \infty

x \to 4 lim (4 - x^{2})

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

x \to - 2 lim (2 x - 3)

x \to 3 lim (+ (x - 2))

n \to \infty lim (cos (2 πn))

What is the limit as x approaches-1 of ln|1+x| ?
The limit as x approaches-1 of ln|1+x| is -infinity