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

The limit as t approaches 0 of (sin^2(3t))/t is 0

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

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

0

Solution steps

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

Apply L'Hopital's Rule

: t \to 0 lim (\frac{3 sin ( 6 t )}{1})

= t \to 0 lim (\frac{3 sin ( 6 t )}{1})

Plug in the value

t = 0

= \frac{3 sin ( 6 \cdot 0 )}{1}

Simplify

\frac{3 sin ( 6 \cdot 0 )}{1} : 0

= 0

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2} + z ^{2}})

\frac{d}{d x} (2 - tan (x))

\int \frac{lo g _{e} ( x )}{x} d x

\frac{\partial}{\partial x} (17 - x^{2} - y^{2})

\frac{d}{d x} (\frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}})

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