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**Prove It** Have you ever heard of L'Hospital's Rule? It states that if you have an indeterminate limit $\displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*}$ or $\displaystyle \displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$, then $\displaystyle \displaystyle \begin{align*} \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \end{align*}$.

Here you have one of the $\displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*}$ form, so L'Hospital's Rule is appropriate.

$\displaystyle \displaystyle \begin{align*} \lim_{t \to 0} \frac{\tan^2{3t}}{2t} &= \lim_{ t \to 0} \frac{\frac{d}{dt}\left( \tan^2{3t} \right) }{\frac{d}{dt} \left(2t\right)} \\ &= \lim_{ t \to 0} \frac{ 6\sec^2{3t}\tan{3t} }{ 2 } \\ &= \frac{6 \sec^2{0} \tan{0}}{2} \\ &= 0 \end{align*}$