limit as x approaches 0+of ln(x)*tan(x)

Lösung

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

0

Schritte zur Lösung

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

Umformung für L'Hopital

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

Wende das L'Hopital Theorem an

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

Vereinfache

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

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

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

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

x \to 0 + lim (1) = 1

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

= - \frac{1}{\infty}

Wende die Unendlichkeitseigenschaft an:

\frac{c}{\infty} = 0

= 0

x \frac{d y}{d x} + 2 y = 2 e^{x}

in v erse l a pl a ce \frac{5}{π s ^{2}}

t an g e n t 8 e^{x}

y^{^{'}} + \frac{3 y}{x} = - \frac{1}{x}

\frac{\partial}{\partial x} (2 x^{3} + x y^{2})