limit as x approaches 0+of tan(5x)ln(8x)

Lösung

x \to 0 + lim (tan (5 x) ln (8 x))

0

Schritte zur Lösung

x \to 0 + lim (tan (5 x) ln (8 x))

Umformung für L'Hopital

= x \to 0 + lim (\frac{ln ( 8 x )}{\frac{1}{t a n ( 5 x )}})

Wende das L'Hopital Theorem an

= x \to 0 + lim (\frac{\frac{1}{x}}{- csc ^{2} ( 5 x ) \cdot 5})

Vereinfache

\frac{\frac{1}{x}}{- csc ^{2} ( 5 x ) \cdot 5} : - \frac{1}{5 x csc ^{2} ( 5 x )}

= x \to 0 + lim (- \frac{1}{5 x csc ^{2} ( 5 x )})

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

= - \frac{1}{5} \cdot x \to 0 + lim (\frac{1}{x csc ^{2} ( 5 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{1}{5} \cdot \frac{lim _{x \to 0 +} ( 1 )}{lim _{x \to 0 +} ( x csc ^{2} ( 5 x ) )}

x \to 0 + lim (1) = 1

x \to 0 + lim (x csc^{2} (5 x)) = \infty

= - \frac{1}{5} \cdot \frac{1}{\infty}

Vereinfache

- \frac{1}{5} \cdot \frac{1}{\infty} : 0

= 0

a re a y = 4 x^{2}, y = 234

x \to \frac{1}{4} lim (tan (π x))

\frac{d}{d x} (\frac{6 x}{x + 3})

\frac{d}{d x} (arccot (x - 1))

x \to \frac{π}{2} lim (\frac{tan ( 2 x )}{x - \frac{π}{2}})