limit as x approaches 0 of (\sqrt[3]{1+x}-1)/(sqrt(1+x)-1)

Lösung

x \to 0 lim (\frac{3 1 + x - 1}{1 + x - 1})

\frac{2}{3}

Dezimale

0.66666 \dots

Schritte zur Lösung

x \to 0 lim (\frac{3 1 + x - 1}{1 + x - 1})

Wende das L'Hopital Theorem an

= x \to 0 lim \frac{\frac{1}{3 ( x + 1 ) ^{\frac{2}{3}}}}{\frac{1}{2 x + 1}}

Setze den Wert

x = 0

ein

= \frac{\frac{1}{3 ( 0 + 1 ) ^{\frac{2}{3}}}}{\frac{1}{2 0 + 1}}

Vereinfache

\frac{\frac{1}{3 ( 0 + 1 ) ^{\frac{2}{3}}}}{\frac{1}{2 0 + 1}} : \frac{2}{3}

= \frac{2}{3}

h \to 0 lim (\frac{2 x + 2 h + 1 - 2 x + 1}{h})

x \to 3 lim (\frac{1}{( 3 + x ) ^{2}})

x \to 7 lim (x + 7)

x \to 1 + lim (\frac{4}{x ^{2} - 1})

x \to 0 - lim (\frac{x}{1 + e ^{\frac{1}{x}}})