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Can anyone help me see how to evaluate the following limit:

$$\lim_{x\to 0} \frac{\tan^2 (3x)}{x^2}$$

Please, tell how can I solve it, I just have entered Calculus I, but if you can don't tell the final answer.

Thank you, :)

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Simply use equivalents: $\;\tan u\sim_0u$, hence $$\frac{\tan^23x}{x^2}\sim_0\frac{(3x)^2}{x^2}=9.$$

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Using the known limit $$ \lim_{x\to0}\frac{\sin t}{t}=1 $$ yields $$ \lim_{x\to0}\frac{\tan^2(3x)}{x^2}= \lim_{x\to0}\left(\frac{\sin(3x)}{3x}\right)^2\left(\frac{3}{\cos(3x)}\right)^2= 1^2\cdot\left(\frac{3}{\cos(0)}\right)^2=9 $$

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Note that $$\frac{\tan^2(3x)}{x^2}=\frac{1}{\cos^2 (3x)}\left(\frac{3\sin 3x}{3x}\right)^2=\frac{9}{\cos^2(3x)}\left(\frac{\sin 3x}{3x}\right)^2$$ and use your knowledge of well-known limits and limit laws namely $$\lim_{u \rightarrow 0}\frac{\sin u}{u}=1.$$

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