Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$

Question

Limit to evaluate: $$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$

Proposed solution: $$ \cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)} \Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/= \frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5} {\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6} $$

Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get $$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$

Please tell me if I am correct.

Answer 1

Hint:

$${f(x)\over g(x)}={{f(x)\over x}\over{g(x)\over x}}$$

Answer 2

Use $$ 2·\cos x·\sin(kx) = \sin((k+1)x)+\sin((k-1)x) $$ or $$ 2\sin(x/2)\sin(kx)=\cos((k-1/2)x)-\cos((k+1/2)x) $$ or something similar.

Answer 3

HINT:

$$\sin (ax) =(ax)+O(x^3)$$

as $x\to 0$.

Answer 4

HINT:

Using Prosthaphaeresis Formula,

$$\sin(a-2)x+\sin(ax)+\sin(a+2)x$$ $$=\sin(ax)+[\sin(a-2)x+\sin(a+2)x]$$ $$=\sin(ax)[1+2\cos2x]$$

Answer 5

The method in the current version of the question is correct.

I suppose to be a bit nitpicking, one might add that in things like $$ \lim_{x\to0} \frac{\sin(3x)}{3x} $$ one should note that as $x\to0$, one also has $3x\to0$. Then one has $$ \lim_{3x\to0} \frac{\sin(3x)}{3x} = \lim_{u\to0}\frac{\sin u} u = 1. $$