limit as x approaches infinity of (x^2)/(e^{4x)}

Lösung

x \to \infty lim (\frac{x ^{2}}{e ^{4 x}})

0

Schritte zur Lösung

x \to \infty lim (\frac{x ^{2}}{e ^{4 x}})

Wende Exponentenregel an:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{2}}{e ^{4 x}} = (x^{2}) e^{- 4 x}

= x \to \infty lim (x^{2} e^{- 4 x})

Umformung für L'Hopital

= x \to \infty lim (\frac{x ^{2}}{\frac{1}{e ^{- 4 x}}})

Wende das L'Hopital Theorem an

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

Wende das L'Hopital Theorem an

= x \to \infty lim (\frac{2}{16 e ^{4 x}})

Streiche die gemeinsamen Faktoren:

2

= x \to \infty lim (\frac{1}{8 e ^{4 x}})

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

= \frac{1}{8} \cdot x \to \infty lim (\frac{1}{e ^{4 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

= \frac{1}{8} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( e ^{4 x} )}

x \to \infty lim (1) = 1

x \to \infty lim (e^{4 x}) = \infty

= \frac{1}{8} \cdot \frac{1}{\infty}

Vereinfache

\frac{1}{8} \cdot \frac{1}{\infty} : 0

= 0

d er i v a t i v e 3 arctan (x + 1 + x^{2})

\int \frac{cot ( x )}{sin ^{2} ( x )} d x

n = 0 \sum \infty x (1 - x)^{n - 1}

\int 1 - 4 x^{2} d x

\frac{d}{d x} ((x^{2} + 1) (x + 1)^{2})