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

Lösung

x \to - 1 + lim (ln ∣ x + 1 ∣)

- \infty

Schritte zur Lösung

x \to - 1 + lim (ln ∣ x + 1 ∣)

x + 1

ist positiv wenn

x \to - 1 +

. Deshalb

∣ x + 1 ∣ = x + 1

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

Wende die Kettenregel für Grenzen an:

- \infty

= - \infty

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

t an g e n t f (x) = \frac{x}{x ^{2} - 15}, at x = 4

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

d er i v a t i v e f (x) = 7 e^{x^{2} - 5 x + 6}

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