limit as x approaches infinity of (2x)/(2^xln(2))

Lösung

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

0

Schritte zur Lösung

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

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

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

Wende das L'Hopital Theorem an

= \frac{2}{ln ( 2 )} \cdot x \to \infty lim (\frac{1}{2 ^{x} ln ( 2 )})

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

= \frac{2}{ln ( 2 )} \cdot \frac{1}{ln ( 2 )} \cdot x \to \infty lim (\frac{1}{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

= \frac{2}{ln ( 2 )} \cdot \frac{1}{ln ( 2 )} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 2 ^{x} )}

x \to \infty lim (1) = 1

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

= \frac{2}{ln ( 2 )} \cdot \frac{1}{ln ( 2 )} \cdot \frac{1}{\infty}

Vereinfache

\frac{2}{ln ( 2 )} \cdot \frac{1}{ln ( 2 )} \cdot \frac{1}{\infty} : 0

= 0

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

x \to 3 lim (x^{2} + 9 x - 27)

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

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

x \to 1 lim (\frac{1 - x}{x})