limit as x approaches 0+of 5sin(x)ln(x)

Lösung

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

0

Schritte zur Lösung

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

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

= 5 \cdot x \to 0 + lim (sin (x) ln (x))

Umformung für L'Hopital

= 5 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{s i n ( x )}})

Wende das L'Hopital Theorem an

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

Vereinfache

\frac{\frac{1}{x}}{- cot ( x ) csc ( x )} : - \frac{1}{x cot ( x ) csc ( x )}

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

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

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

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

x \to 0 + lim (1) = 1

x \to 0 + lim (x cot (x) csc (x)) = \infty

= 5 (- \frac{1}{\infty})

Vereinfache

5 (- \frac{1}{\infty}) : 0

= 0

\frac{\partial}{\partial x} (7 e^{x^{7} y})

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{6 x ^{\frac{5}{6}}})

\frac{\partial}{\partial x} (\frac{y + x y}{( 1 + x ) ^{3}})

\frac{d}{d x} (e^{5 x} \cdot 5)

d er i v a t i v e u + 1 (1 - 8 u^{2})^{7}