Can anybody help me find this limit without using L'Hospital's rule?
$$\lim_{x \rightarrow 0} \, \frac{\arctan(3x)}{\tan\big((x+3\pi)/3\big)}$$
I've tried to multiply both $\arctan/\tan$ on $\sin$, but it doesn't seem to help.
Answer
Use the following facts.
$\displaystyle \lim_{x \to 0} \frac{\arctan(3x)}{\tan((x+3\pi)/3)} = \lim_{x \to 0} \frac{\arctan(3x)}{3x} \times \frac{3x}{\tan\frac{x+3\pi}{3}}$
$\displaystyle\tan\left(\pi +\frac{x}{3}\right) = \tan\left(\frac{x}{3}\right)$
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