# How do you find the limit of sin^2(3t)/t^2 as t->0?

Nov 9, 2016

${\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = 9$

#### Explanation:

graph{((sin(3x))/x) * ((sin(3x))/x) [-9 9, -0.5, 10.5]}

${\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = {\lim}_{t \rightarrow 0} {\left(\sin \frac{3 t}{t}\right)}^{2}$
$\therefore {\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = {\left({\lim}_{t \rightarrow 0} \sin \frac{3 t}{t}\right)}^{2}$
$\therefore {\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = {\left({\lim}_{t \rightarrow 0} \frac{3}{3} \sin \frac{3 t}{t}\right)}^{2}$
$\therefore {\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = {\left(3 {\lim}_{t \rightarrow 0} \frac{1}{3} \sin \frac{3 t}{t}\right)}^{2}$
$\therefore {\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = {\left(3\right)}^{2} {\left({\lim}_{t \rightarrow 0} \sin \frac{3 t}{3 t}\right)}^{2}$
$\therefore {\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = 9 {\left({\lim}_{t \rightarrow 0} \sin \frac{3 t}{3 t}\right)}^{2}$

Let $\theta = 3 t$, Then As $t \rightarrow 0 \implies \theta \to 0$, So:

$\therefore {\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = 9 {\left({\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta}\right)}^{2}$

And, ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$is a standard calculus limit.

Hence,

$\therefore {\lim}_{t \rightarrow 0} {\sin}^{2} \frac{3 t}{t} ^ 2 = 9 {\left(1\right)}^{2} = 9$

Which is consistent with the above graph.