What is the limit as t approaches-1 of ln(t+2) ?

The limit as t approaches-1 of ln(t+2) is 0

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limit as t approaches-1 of ln(t+2)

t \to - 1 lim (ln (t + 2))

0

Solution steps

t \to - 1 lim (ln (t + 2))

Plug in the value

t = - 1

= ln (- 1 + 2)

Simplify

ln (- 1 + 2) : 0

= 0

y = \frac{z ^{2} + 1}{z ^{2} - 1}

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

\frac{d}{d x} (\frac{d y}{d x}) + y = - 1

\int (e^{x} + e) d x

\frac{\partial}{\partial x} (- (x - 2) (y - 2) (x + y - 3))

What is the limit as t approaches-1 of ln(t+2) ?
The limit as t approaches-1 of ln(t+2) is 0