What is the limit as x approaches 0+of 2xln(sin(x)) ?

The limit as x approaches 0+of 2xln(sin(x)) is 0

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limit as x approaches 0+of 2xln(sin(x))

x \to 0 + lim (2 x ln (sin (x)))

0

Solution steps

x \to 0 + lim (2 x ln (sin (x)))

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

= 2 \cdot x \to 0 + lim (x ln (sin (x)))

Rewrite for L'Hopital

= 2 \cdot x \to 0 + lim (\frac{ln ( sin ( x ) )}{\frac{1}{x}})

Apply L'Hopital's Rule

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

Simplify

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

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

Plug in the value

x = 0

= 2 (- \frac{2 \cdot 0}{sec ^{2} ( 0 )})

Simplify

2 (- \frac{2 \cdot 0}{sec ^{2} ( 0 )}) : 0

= 0

What is the limit as x approaches 0+of 2xln(sin(x)) ?
The limit as x approaches 0+of 2xln(sin(x)) is 0