Find limit $\lim_{x \rightarrow 0} \, \frac{\arctan(3x)}{\tan\big((x+3\pi)/3\big)}$ without using L'Hospital's rule

Question

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 1

Use the following facts.

• $\displaystyle \lim_{x \to 0} \frac{\arctan(3x)}{\tan((x+3\pi)/3)} = \lim_{x \to 0} \frac{\arctan(3x)}{3x} \times \frac{3x}{\tan\frac{x+3\pi}{3}}$

• $\displaystyle\tan\left(\pi +\frac{x}{3}\right) = \tan\left(\frac{x}{3}\right)$

Answer 2

We have

$$\lim_{x\to0}\frac{\arctan(3x)}{\tan ((x+3\pi)/3)}=\lim_{x\to0}\frac{\arctan(3x)}{\tan (x/3)}=\lim_{x\to0}3\frac{x/3}{\tan (x/3)}\frac{\arctan (3x)}{x}\\=9\arctan'(0)=\frac9{1+0^2}=9$$

Answer 3

Since $$\operatorname{tg}{\frac{x+3\pi}{3}}=\operatorname{tg}{\left(\frac{x}{3}+\pi\right)}=\operatorname{tg}{\left(\frac{x}{3}\right)} \underset{x\to{0}}\sim \frac{x}{3},\\ \operatorname{arctg}{(3x)}\underset{x\to{0}}\sim 3x,$$ then $$\dfrac{\operatorname{arctg}{(3x)}}{\operatorname{tg}{\frac{x+3\pi}{3}}}\underset{x\to{0}}\to 9.$$