1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Use the following facts.

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

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

$\endgroup$
1
$\begingroup$

We have

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

$\endgroup$
0
$\begingroup$

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 $$\dfrac{\operatorname{arctg}{(3x)}}{\operatorname{tg}{\frac{x+3\pi}{3}}}\underset{x\to{0}}\to 9.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.