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

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

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limit as x approaches 0+of xln(tan(x))

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= - 2 \cdot \frac{2 \cdot 0}{cos ( 2 \cdot 0 ) \cdot 2}

Simplify

- 2 \cdot \frac{2 \cdot 0}{cos ( 2 \cdot 0 ) \cdot 2} : 0

= 0

x \to 2 - lim (\frac{- x}{x - 2})

x \to - 6 lim (\frac{∣ x ∣}{x})

x \to 1 + lim (\frac{e ^{x} + 1}{1 - x})

x \to 0 lim ((\frac{1 + 5 x}{1 + 2 x})^{\frac{1}{x}})

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

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