What is the limit as x approaches 0 of (xtan(2x))/(sin^2(3x)) ?

The limit as x approaches 0 of (xtan(2x))/(sin^2(3x)) is 2/9

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limit as x approaches 0 of (xtan(2x))/(sin^2(3x))

x \to 0 lim (\frac{x tan ( 2 x )}{sin ^{2} ( 3 x )})

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

x \to 0 lim (\frac{x tan ( 2 x )}{sin ^{2} ( 3 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{tan ( 2 x ) + 2 x sec ^{2} ( 2 x )}{3 sin ( 6 x )})

= x \to 0 lim (\frac{tan ( 2 x ) + 2 x sec ^{2} ( 2 x )}{3 sin ( 6 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{4 sec ^{2} ( 2 x ) + 8 x sec ^{2} ( 2 x ) tan ( 2 x )}{18 cos ( 6 x )})

= x \to 0 lim (\frac{4 sec ^{2} ( 2 x ) + 8 x sec ^{2} ( 2 x ) tan ( 2 x )}{18 cos ( 6 x )})

Plug in the value

x = 0

= \frac{4 sec ^{2} ( 2 \cdot 0 ) + 8 \cdot 0 \cdot sec ^{2} ( 2 \cdot 0 ) tan ( 2 \cdot 0 )}{18 cos ( 6 \cdot 0 )}

Simplify

\frac{4 sec ^{2} ( 2 \cdot 0 ) + 8 \cdot 0 \cdot sec ^{2} ( 2 \cdot 0 ) tan ( 2 \cdot 0 )}{18 cos ( 6 \cdot 0 )} : \frac{2}{9}

= \frac{2}{9}

x \to 3 lim (x^{2} + 7 x + 2)

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

x \to \infty lim (3 x + 5)

m \to 0 lim (\frac{sin ( m )}{m})

x \to 1 - lim (\frac{arccos ( x )}{1 - x ^{2}})

What is the limit as x approaches 0 of (xtan(2x))/(sin^2(3x)) ?
The limit as x approaches 0 of (xtan(2x))/(sin^2(3x)) is 2/9