$\lim\limits_{x \to 0} \frac{\tan(mx)}{\tan(nx)} =$?

Question

I've tried by writing it as $\tan(m x) \cot(n x)$, but I couldn't really continue from there.

Answer 1

Hint: ${{\tan(mx)}\over{\tan(nx)}}=m{{\sin(mx)}\over{mx}}{1\over n}{{(nx)}\over{\sin(nx)}}{{\cos(nx)}\over{\cos(mx)}}$

so the limit is ${m\over n}$.

Answer 2

$$\tan(mx) \cot(nx)=\frac{\cos(nx)\sin(mx)}{\cos(mx)\sin(nx)}$$

First-order Taylor expansion around zero gives $$\cos(nx)=1+O(x^2)$$$$\cos(mx)=1+O(x^2)$$$$\sin(nx)=nx+O(x^3)$$$$\sin(mx)=mx+O(x^3)$$

therefore $$\lim_{x \rightarrow 0}{\tan(mx) \cot(nx)}=\lim_{x \rightarrow 0}\frac{mx}{nx}=\frac{m}{n}$$

Answer 3

Hint: Prove $$\lim\limits_{x \to 0} \frac{\tan(x)}{x} =1$$ using $$\lim\limits_{x \to 0} \frac{\sin(x)}{x} =1$$

Answer 4

If you know $\lim_{x \to 0} \frac{\sin(x)}{x}=1$ expand it to

$$\lim_{x \to 0}\frac{\tan(mx)}{\tan(nx)} = \lim_{x \to 0}\frac{\sin(mx)}{mx} \cdot \frac{mx}{\cos(mx)} \cdot \frac{\cos(nx)}{nx} \cdot \frac{nx}{\sin(nx)}$$

Answer 5

The shortest way, as often, uses asymptotic equivalents: near $0$, $\tan u\sim u $, hence $$\frac{\tan mx}{\tan nx}\sim_0\frac{mx}{nx}=\frac mn.$$