limit as x approaches 0+of x[ln(x)]^2

Lösung

x \to 0 + lim (x [ln (x)]^{2})

0

Schritte zur Lösung

x \to 0 + lim (x [ln (x)]^{2})

Vereinfache

= x \to 0 + lim (x ln^{2} (x))

Umformung für L'Hopital

= x \to 0 + lim (\frac{ln ^{2} ( x )}{\frac{1}{x}})

Wende das L'Hopital Theorem an

= x \to 0 + lim (\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}})

Vereinfache

\frac{\frac{2 l n ( x )}{x}}{- \frac{1}{x ^{2}}} : - 2 x ln (x)

= x \to 0 + lim (- 2 x ln (x))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= - 2 \cdot x \to 0 + lim (x ln (x))

Umformung für L'Hopital

= - 2 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x}})

Wende das L'Hopital Theorem an

= - 2 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})

Vereinfache

\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}} : - x

= - 2 \cdot x \to 0 + lim (- x)

Setze den Wert

x = 0

ein

= - 2 (- 0)

Vereinfache

= 0

\frac{d}{d x} (2^{5 x} 3^{4 x^{2}})

\frac{d y}{d x} = \frac{x - 4}{y + 5}

t an g e n t f (x) = \frac{2}{x}, (6, \frac{1}{3})

\int 15 ln (x + 5) d x

d er i v a t i v e 7 e^{x} + \frac{8}{3 x}