limit as x approaches 1 of (x^{2n}-2x^n+1)/(x^{2m)-2x^m+1}

Lösung

x \to 1 lim (\frac{x ^{2 n} - 2 x ^{n} + 1}{x ^{2 m} - 2 x ^{m} + 1})

\frac{n ^{2}}{m ^{2}}

Schritte zur Lösung

x \to 1 lim (\frac{x ^{2 n} - 2 x ^{n} + 1}{x ^{2 m} - 2 x ^{m} + 1})

Wende das L'Hopital Theorem an

= x \to 1 lim (\frac{2 n x ^{2 n - 1} - 2 n x ^{n - 1}}{2 m x ^{2 m - 1} - 2 m x ^{m - 1}})

Wende das L'Hopital Theorem an

= x \to 1 lim (\frac{2 n x ^{2 n - 2} ( 2 n - 1 ) - 2 n x ^{n - 2} ( n - 1 )}{2 m x ^{2 m - 2} ( 2 m - 1 ) - 2 m x ^{m - 2} ( m - 1 )})

Setze den Wert

x = 1

ein

= \frac{2 n \cdot 1 ^{2 n - 2} ( 2 n - 1 ) - 2 n \cdot 1 ^{n - 2} ( n - 1 )}{2 m \cdot 1 ^{2 m - 2} ( 2 m - 1 ) - 2 m \cdot 1 ^{m - 2} ( m - 1 )}

Vereinfache

\frac{2 n \cdot 1 ^{2 n - 2} ( 2 n - 1 ) - 2 n \cdot 1 ^{n - 2} ( n - 1 )}{2 m \cdot 1 ^{2 m - 2} ( 2 m - 1 ) - 2 m \cdot 1 ^{m - 2} ( m - 1 )} : \frac{n ^{2}}{m ^{2}}

= \frac{n ^{2}}{m ^{2}}

\frac{d P}{d t} = Pt, P (1) = 2

\int_{0}^{\frac{π}{4}} tan^{4} (t) d t

\frac{d}{d x} (a - b x)

\int - \frac{sin ( x )}{cos ^{3} ( x )} d x

\frac{d}{d x} ((sin (x) + x)^{2} + 7 cos (5 x))