limit as x approaches 0+of xln(tan(5x))

Lösung

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

0

Schritte zur Lösung

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

Umformung für L'Hopital

= x \to 0 + lim (\frac{ln ( tan ( 5 x ) )}{\frac{1}{x}})

Wende das L'Hopital Theorem an

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

Vereinfache

= x \to 0 + lim (- 5 x^{2} sec^{2} (5 x) cot (5 x))

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

= - 5 \cdot x \to 0 + lim (x^{2} sec^{2} (5 x) cot (5 x))

Drücke mit sin, cos aus

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

\frac{x ^{2}}{cos ( 5 x ) sin ( 5 x )} = \frac{x ^{2}}{sin ( 5 x )} \cdot \frac{1}{cos ( 5 x )}

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

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

mit Einschränkung des unbestimmten Ausdrucks

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

x \to 0 + lim (\frac{x ^{2}}{sin ( 5 x )}) = 0

x \to 0 + lim (\frac{1}{cos ( 5 x )}) = 1

= - 5 \cdot 0 \cdot 1

Vereinfache

= 0

in v erse l a pl a ce 4 s

t an g e n t f (x) = \frac{2}{x}, at x = - 9

\frac{d}{d x} ((x^{2} + 2 x) e^{x})

t an g e n t x^{2} + 2 x y - y^{2} + x = 51, (5, 7)

s l o p e (4.3) (4.6)