Evaluate $\lim_{x \to 0}\frac{\sin(mx)}{\sin(nx)}$ without L'Hospital theorem

Question

The limit

$$\lim_{x \to 0}\frac{\sin(mx)}{\sin(nx)}$$

can be easily found through the L'Hospital theorem.

However I would like to know how to reach the result without employing it.

I have tried using Euler's identity and the multiple angle formulas, but I'm unable to reach the result.

Answer 1

HINT

\begin{align*} \lim_{x\to 0}\frac{\sin(mx)}{\sin(nx)} = \lim_{x\to 0}\left(\frac{\sin(mx)}{mx}\times\frac{nx}{\sin(nx)}\times\frac{m}{n}\right) \end{align*}

Answer 2

As a matter a fact, I would not recommend using l'Hospital rule for computing this limit. L'Hospital involves the derivative of $\text{sin}$ function. How do we show that $\text{sin}' = \text{cos}$? Well, using the fundamenatal limit $$\lim_{x \rightarrow \ 0} \frac{\text{sin}(x)}{x} = 1$$ This is actually what you need for computing your limit.

Answer 3

As you've said you couldn't get result from Euler's identity then well let's do it ;)
Rewriting using Euler's identity we get $$L=\lim_{x\to 0} \frac{e^{mx}-e^{-mx}}{e^{nx}-e^{-nx}}$$ $$L=\lim_{x\to 0} \frac{e^{2mx}-1}{e^{2nx}-1}\cdot \frac{e^{nx}}{e^{mx}} = \lim_{x\to 0} \frac{e^{2mx}-1}{e^{2nx}-1}\cdot \underbrace{\lim_{x\to 0} e^{(n-m)x}}_{=1}$$ Now the Taylor series of the exponential function gives : $$e^x=1+x+\frac{x^2}{2!}+..+..+\ldots $$ Hence we have : $$e^{2mx}=1+(2mx)+\frac{4m^2x^2}{2!}+..+..+\ldots$$ Transposing terms and rearranging we have : $$\frac{e^{2mx}-1}{x}=2m+\frac{4m^2x}{2!}+..+..+\ldots $$ Letting $x$ approach 0 gives $$ \lim_{x\to 0} \frac{e^{2mx}-1}{x}=2m$$ Similarly we have $$ \lim_{x\to 0} \frac{e^{2nx}-1}{x}=2n$$ Dividing these two equation gives : $$L=\lim_{x\to 0} \frac{e^{2mx}-1}{e^{2nx}-1} = \frac{2m}{2n}=\frac{m}{n}$$ Hence the answer. ;)

Answer 4

Still simpler with some basic asymptotic analysis: we know that near $0$, $\sin u$ is equivalent to $u$, and equivalence of functions is compatible with multiplication and division, so $$\frac{\sin mx}{\sin nx}\sim_0\frac{mx}{nx}=\frac mn.$$