limit as x approaches 0+of sin(x)ln(4x)

Lösung

x \to 0 + lim (sin (x) ln (4 x))

0

Schritte zur Lösung

x \to 0 + lim (sin (x) ln (4 x))

Wende die folgende algebraische Eigenschaft an

: a \cdot b = \frac{a}{\frac{1}{b}}, b \neq = 0

ln (4 x) sin (x) = \frac{ln ( 4 x )}{\frac{1}{s i n ( x )}}

= x \to 0 + lim (\frac{ln ( 4 x )}{\frac{1}{s i n ( x )}})

Wende das L'Hopital Theorem an

= 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 )}

= 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)

= - 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

= - \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

= - \frac{1}{\infty}

Wende die Unendlichkeitseigenschaft an:

\frac{c}{\infty} = 0

= 0

\int t cos (n t) d t

k \to \infty lim (\frac{k ^{\frac{1}{k}}}{ln ( k )})

\frac{d}{d s} (\frac{s ^{2} - 4 s + 20}{( s + 2 ) ( s + 4 )})

\int e^{t a n (6 x)} sec^{2} (6 x) d x

y^{^{'}} = 6 x^{2} y