Answer
$\frac{1}{8}$
Work Step by Step
First, simplify as much as possible.
$=\lim\limits_{t \to 0}\frac{t^3}{\tan^3(2t)}$
$=\left(\lim\limits_{t \to 0}\frac{t}{\tan(2t)}\right)^3$
Since $t \to 0$, let us linearly approximate $f(t) = \tan(t)$ at $a=0$.
$L(t) = f(a) + f'(a)(t-a)$
$L(t) = \tan(0) + \sec^2(0)(t-0)$
$L(t) = t$
Thus, as $t$ is near 0, $\tan(t) \approx t$. Additionally, at $t=0$, $\tan(t)=t$. We have $\tan(2t)$ in the denominator of the limit, but $2t=t=0$. So, let us continue solving the limit using $\tan(2t)=2t$.
$=(\lim\limits_{t \to 0}\frac{t}{\tan(2t)})^3$
$=(\lim\limits_{t \to 0}\frac{t}{2t})^3$
$=(\lim\limits_{t \to 0}\frac{1}{2})^3$
$=(\frac{1}{2})^3$
$=\frac{1}{8}$
