limit as x approaches 0 of ((1-cos^2(x)))/(ln(1+x^2))

Lösung

x \to 0 lim (\frac{( 1 - cos ^{2} ( x ) )}{ln ( 1 + x ^{2} )})

1

Schritte zur Lösung

x \to 0 lim (\frac{1 - cos ^{2} ( x )}{ln ( 1 + x ^{2} )})

Vereinfache

\frac{1 - cos ^{2} ( x )}{ln ( 1 + x ^{2} )} : \frac{sin ^{2} ( x )}{ln ( 1 + x ^{2} )}

= x \to 0 lim (\frac{sin ^{2} ( x )}{ln ( 1 + x ^{2} )})

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{sin ( 2 x )}{\frac{2 x}{1 + x ^{2}}})

Vereinfache

= x \to 0 lim (\frac{sin ( 2 x ) ( 1 + x ^{2} )}{2 x})

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{cos ( 2 x ) \cdot 2 ( 1 + x ^{2} ) + 2 x sin ( 2 x )}{2})

Setze den Wert

x = 0

ein

= \frac{cos ( 2 \cdot 0 ) \cdot 2 ( 1 + 0 ^{2} ) + 2 \cdot 0 \cdot sin ( 2 \cdot 0 )}{2}

Vereinfache

\frac{cos ( 2 \cdot 0 ) \cdot 2 ( 1 + 0 ^{2} ) + 2 \cdot 0 \cdot sin ( 2 \cdot 0 )}{2} : 1

= 1

x \to 0 lim (x^{9} e^{- x^{8}})

x \to 1 lim (2 x + 2)

x \to 0 + lim (\frac{x}{x ^{2} + 1})

x \to \infty lim (\frac{1}{6} ln (tanh (\frac{3 x}{2})))

x \to 1 lim (\frac{2 x - 2}{3 26 + x - 3})