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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{2 x}{sec ^{2} ( 2 x ) \cdot 2})

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

Plug in the value

x = 0

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

Simplify

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

= 0

x \to 5 lim (\frac{0}{0})

x \to o - lim (f (x))

x \to 10 lim (7)

x \to 0 - lim (tan (x) + \frac{π}{2})

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