limit as x approaches infinity+of x^{(3)}e^{(-x^{(2)})}

Lösung

x \to \infty + lim (x^{(3)} e^{(- x^{(2)})})

0

Schritte zur Lösung

x \to \infty lim (x^{3} e^{- x^{2}})

Umformung für L'Hopital

= x \to \infty lim (\frac{x ^{3}}{\frac{1}{e ^{- x^{2}}}})

Wende das L'Hopital Theorem an

= x \to \infty lim (\frac{3 x ^{2}}{e ^{x^{2}} \cdot 2 x})

Wende das L'Hopital Theorem an

= x \to \infty lim (\frac{6 x}{2 ( 2 e ^{x^{2}} x ^{2} + e ^{x^{2}} )})

Wende das L'Hopital Theorem an

= x \to \infty lim (\frac{6}{2 ( 4 e ^{x^{2}} x ^{3} + 6 e ^{x^{2}} x )})

Vereinfache

= x \to \infty lim (\frac{3}{4 e ^{x^{2}} x ^{3} + 6 e ^{x^{2}} x})

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

= 3 \cdot x \to \infty lim (\frac{1}{4 e ^{x^{2}} x ^{3} + 6 e ^{x^{2}} x})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

mit Einschränkung des unbestimmten Ausdrucks

= 3 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 4 e ^{x^{2}} x ^{3} + 6 e ^{x^{2}} x )}

x \to \infty lim (1) = 1

x \to \infty lim (4 e^{x^{2}} x^{3} + 6 e^{x^{2}} x) = \infty

= 3 \cdot \frac{1}{\infty}

Vereinfache

3 \cdot \frac{1}{\infty} : 0

= 0

\int x^{11} e^{x^{12}} d x

\int x (2 + x^{2}) d x

n = 1 \sum \infty 4^{- n}

x \to 4 + lim (e^{\frac{7}{( 4 - x )}})

d er i v a t i v e 4 sin (x) cos (x)