limit as x approaches infinity of (e^{5x-3})/(ln(x-2))

Lösung

x \to \infty lim (\frac{e ^{5 x - 3}}{ln ( x - 2 )})

\infty

Schritte zur Lösung

x \to \infty lim (\frac{e ^{5 x - 3}}{ln ( x - 2 )})

Wende das L'Hopital Theorem an

= x \to \infty lim (\frac{e ^{5 x - 3} \cdot 5}{\frac{1}{x - 2}})

Vereinfache

= x \to \infty lim (5 e^{5 x - 3} (x - 2))

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

= 5 \cdot x \to \infty lim (e^{5 x - 3} (x - 2))

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

mit Einschränkung des unbestimmten Ausdrucks

= 5 \cdot x \to \infty lim (e^{5 x - 3}) \cdot x \to \infty lim (x - 2)

x \to \infty lim (e^{5 x - 3}) = \infty

x \to \infty lim (x - 2) = \infty

= 5 \cdot \infty \cdot \infty

Wende die Unendlichkeitseigenschaft an:

5\infty \infty = \infty

= \infty

d er i v a t i v e f (x) = cos (a^{8} + x^{8})

\int_{0}^{\infty} x^{1} e^{- x} d x

\frac{d}{d x} (ln ((5 x^{2} + 9)^{3}))

\int (sin (2 x) - cos (2 x))^{2} d x

\frac{\partial}{\partial x} (e^{6 x e^{y}})