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

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

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

x \to - 1 + lim (ln ∣ x + 1 ∣)

- \infty

Solution steps

x \to - 1 + lim (ln ∣ x + 1 ∣)

x + 1

is positive when

x \to - 1 +

. Therefore

∣ x + 1 ∣ = x + 1

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

Apply the Limit Chain Rule:

- \infty

= - \infty

\frac{d}{d x} (x^{2} e^{\frac{- x}{10}})

f (x) = \frac{x}{x ^{2} - 15}, at x = 4

y^{^{''}} + 25 y^{^{'}} = 0

f (x) = 7 e^{x^{2} - 5 x + 6}

\int (10 x + 4) 5 x^{2} + 4 x d x

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