limit as n approaches infinity of (n^{sqrt(2)})/(21^n)

Lösung

n \to \infty lim (\frac{n ^{2}}{2 1 ^{n}})

0

Schritte zur Lösung

n \to \infty lim (\frac{n ^{2}}{2 1 ^{n}})

Wende das L'Hopital Theorem an

= n \to \infty lim (\frac{2 n ^{2 - 1}}{2 1 ^{n} ln ( 21 )})

Wende das L'Hopital Theorem an

= n \to \infty lim (\frac{2 ( 2 - 1 ) n ^{2 - 2}}{ln ^{2} ( 21 ) \cdot 2 1 ^{n}})

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

= \frac{2 ( 2 - 1 )}{ln ^{2} ( 21 )} \cdot n \to \infty lim (\frac{n ^{2 - 2}}{2 1 ^{n}})

Multipliziere aus

2 (2 - 1) : 2 - 2

= \frac{2 - 2}{ln ^{2} ( 21 )} \cdot n \to \infty lim (\frac{n ^{2 - 2}}{2 1 ^{n}})

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 - 2}{ln ^{2} ( 21 )} \cdot \frac{lim _{n \to \infty} ( n ^{2 - 2} )}{lim _{n \to \infty} ( 2 1 ^{n} )}

n \to \infty lim (n^{2 - 2}) = 0

n \to \infty lim (2 1^{n}) = \infty

= \frac{2 - 2}{ln ^{2} ( 21 )} \cdot \frac{0}{\infty}

Vereinfache

\frac{2 - 2}{ln ^{2} ( 21 )} \cdot \frac{0}{\infty} : 0

= 0

l a pl a ce t r an s f or m 5 t^{2} - 2 e^{t}

\frac{d}{d x} (- \frac{k}{x ^{2}})

d er i v a t i v e x^{3} - 8 x^{2} + 19 x - 12

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\int \frac{1}{t ^{3} + t ^{2} - 2 t} d t