What is the limit as x approaches 0+of sin(3x)ln(x) ?

The limit as x approaches 0+of sin(3x)ln(x) is 0

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limit as x approaches 0+of sin(3x)ln(x)

x \to 0 + lim (sin (3 x) ln (x))

0

Solution steps

x \to 0 + lim (sin (3 x) ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{s i n ( 3 x )}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{- cot ( 3 x ) csc ( 3 x ) \cdot 3})

= x \to 0 + lim (\frac{\frac{1}{x}}{- cot ( 3 x ) csc ( 3 x ) \cdot 3})

Simplify

\frac{\frac{1}{x}}{- cot ( 3 x ) csc ( 3 x ) \cdot 3} : - \frac{1}{3 x cot ( 3 x ) csc ( 3 x )}

= x \to 0 + lim (- \frac{1}{3 x cot ( 3 x ) csc ( 3 x )})

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

= - \frac{1}{3} \cdot x \to 0 + lim (\frac{1}{x cot ( 3 x ) csc ( 3 x )})

Express with sin, cos

= - \frac{1}{3} \cdot x \to 0 + lim (\frac{sin ^{2} ( 3 x )}{x cos ( 3 x )})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{3 sin ( 6 x )}{cos ( 3 x ) - 3 x sin ( 3 x )})

= - \frac{1}{3} \cdot x \to 0 + lim (\frac{3 sin ( 6 x )}{cos ( 3 x ) - 3 x sin ( 3 x )})

Plug in the value

x = 0

= - \frac{1}{3} \cdot \frac{3 sin ( 6 \cdot 0 )}{cos ( 3 \cdot 0 ) - 3 \cdot 0 \cdot sin ( 3 \cdot 0 )}

Simplify

- \frac{1}{3} \cdot \frac{3 sin ( 6 \cdot 0 )}{cos ( 3 \cdot 0 ) - 3 \cdot 0 \cdot sin ( 3 \cdot 0 )} : 0

= 0

\int \frac{1}{2} u d u

\int_{- 7}^{- 3} \frac{1}{x} d x

\int - e^{x} cos (y) d y

\frac{\partial}{\partial x} (4 x + 6 y + 2 z)

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} ln (t)

What is the limit as x approaches 0+of sin(3x)ln(x) ?
The limit as x approaches 0+of sin(3x)ln(x) is 0