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

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

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limit as x approaches 0 of (sin^2(9x))/x

x \to 0 lim (\frac{sin ^{2} ( 9 x )}{x})

0

Solution steps

x \to 0 lim (\frac{sin ^{2} ( 9 x )}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{18 sin ( 9 x ) cos ( 9 x )}{1})

= x \to 0 lim (\frac{18 sin ( 9 x ) cos ( 9 x )}{1})

Plug in the value

x = 0

= \frac{18 sin ( 9 \cdot 0 ) cos ( 9 \cdot 0 )}{1}

Simplify

\frac{18 sin ( 9 \cdot 0 ) cos ( 9 \cdot 0 )}{1} : 0

= 0

x \to 0 - lim (x^{2} ln (- x))

x \to 0 + lim (x^{3} e^{\frac{1}{x}})

x \to - 1 lim (\frac{x + 1}{x ^{3} - x})

a \to 0 lim (cot (a) - csc (a))

x \to - 1 lim (\frac{3 x ^{2}}{2 x - 1})

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