Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 46

Determine $\displaystyle \lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t}$


$\begin{equation} \begin{aligned} \lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= \lim\limits_{t \to 0} \frac{\frac{t^3}{\sin^3 2t} }{\cos^3 2t} = \lim\limits_{t \to 0} \frac{t^3}{\sin^3 2t} \cdot \cos^3 2t && \Longleftarrow \text{If we introduce } \frac{8}{8} \text{ such that...}\\ \\ \lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= \lim_{t \to 0} \frac{8t^3}{\sin^3 2t} \cdot \frac{\cos^3 2t}{8} && \Longleftarrow \text{we can rewrite the limit as...}\\ \\ \lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= \lim_{t \to 0} \left( \frac{2t}{\sin 2t} \right)^3 \cdot \frac{\cos^3 2t}{8} && \Longleftarrow \text{recall that } \lim_{t \to 0} \frac{\sin t}{t} = 1\\ \\ \lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= \lim_{t \to 0} (1)^3 \cdot \lim_{t \to 0} \frac{\cos^3 2(0)}{8}\\ \\ \lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= 1 \cdot \frac{1}{8}\\ \\ \lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= \frac{1}{8} \end{aligned} \end{equation}$