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

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


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\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...}\\
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\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...}\\
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\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\\
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\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}\\
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\lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= 1 \cdot \frac{1}{8}\\
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\lim\limits_{t \to 0} \frac{t^3}{\tan^3 2t} &= \frac{1}{8}

\end{aligned}
\end{equation}
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