# How do you find the limit of (sin^2(4t)) / t^2  as t approaches 0?

May 16, 2016

Use algebra, properties of limits and the important limit ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$

#### Explanation:

${\sin}^{2} \frac{4 t}{t} ^ 2 = {\left(\sin \frac{4 t}{t}\right)}^{2}$ (Algebra)

So we need only find ${\lim}_{t \rightarrow 0} \sin \frac{4 t}{t}$ and square that result. (Property of limits)

There are 3 big tricks in mathematics:

1. Add $0$
2. Multiply by $1$
3. Change the problem into something we know how
to solve

In this case we'll use item 2. to accomplish 3.

$\sin \frac{4 t}{t} = \frac{4}{4} \sin \frac{4 t}{t} = 4 \left(\sin \frac{4 t}{\left(4 t\right)}\right)$

Now, as $t \rightarrow 0$, we also see that $4 t \rightarrow 0$, so thinking of $4 t$ as $x$ in ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$,

we arrive at ${\lim}_{t \rightarrow 0} \left(\sin \frac{4 t}{\left(4 t\right)}\right) = 1$

So ${\lim}_{t \rightarrow 0} 4 \left(\sin \frac{4 t}{\left(4 t\right)}\right) = 4 \left(1\right) = 4$

And finally

${\left({\lim}_{t \rightarrow 0} \sin \frac{4 t}{t}\right)}^{2} = {\left({\lim}_{t \rightarrow 0} 4 \left(\sin \frac{4 t}{4} t\right)\right)}^{2}$

$= {\left(4\right)}^{2} = 16$