limit as n approaches infinity of n/6 sin(6/n)

Lösung

n \to \infty lim (\frac{n}{6} sin (\frac{6}{n}))

1

Schritte zur Lösung

n \to \infty lim (\frac{n}{6} sin (\frac{6}{n}))

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

= \frac{1}{6} \cdot n \to \infty lim (n sin (\frac{6}{n}))

Umformung für L'Hopital

= \frac{1}{6} \cdot n \to \infty lim (\frac{sin ( \frac{6}{n} )}{\frac{1}{n}})

Wende das L'Hopital Theorem an

= \frac{1}{6} \cdot n \to \infty lim \frac{- \frac{6 c o s ( \frac{6}{n} )}{n ^{2}}}{- \frac{1}{n ^{2}}}

Vereinfache

\frac{- \frac{6 c o s ( \frac{6}{n} )}{n ^{2}}}{- \frac{1}{n ^{2}}} : 6 cos (\frac{6}{n})

= \frac{1}{6} \cdot n \to \infty lim (6 cos (\frac{6}{n}))

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

= \frac{1}{6} \cdot 6 \cdot n \to \infty lim (cos (\frac{6}{n}))

Wende die Kettenregel für Grenzen an:

1

= \frac{1}{6} \cdot 6 \cdot 1

Vereinfache

\frac{1}{6} \cdot 6 \cdot 1 : 1

= 1

\int e^{t a n (5 x)} sec^{2} (5 x) d x

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

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

\frac{\partial}{\partial x} (8 x^{5} y^{6} + 6 x^{4} y^{8})

d er i v a t i v e 6 x^{- 4}