Finding $\lim_{x \to 0} \sin x/\sin(7x)$ without l'Hopital's Rule

Question

$$\lim_{x \to 0} \frac{\sin x}{\sin(7x)}$$

What I did to compute this limit is use $\sin(A+B) = \sin(A)\cos(B) + \cos(B)\sin(A)$ and $\sin(2A) = 2\sin A\cos A$ repeatedly on $\sin(7x)$:

$$ \begin{align} \sin(7x) &= \sin(6x + x) \\ &= \sin(6x)\cos(x) + \cos(6x)\sin(x) \\ &= (2 \sin(3x)\cos(3x))\cos(x) + \cos(6x)\sin(x) \\ &= \big[2 \sin(x + 2x)\cos(3x)\big]\cos(x) + \cos(6x)\sin(x) \\ &= \Big[2 \Big(\sin (x)\cos(2x) + \cos(x)\sin(2x)\Big)\cos(3x)\Big]\cos(x) + \cos(6x)\sin(x) \\ &= \Big[2 \Big(\sin (x)\cos(2x) + 2\cos^2(x)\sin(x)\Big)\cos(3x)\Big]\cos(x) + \cos(6x)\sin(x) \\ &= 2\sin (x) \Big(\cos(2x) + 2\cos^2(x)\Big)\cos(3x)\cos(x) + \cos(6x)\sin(x) \\ &= \sin x \Big[2 \Big(\cos(2x) + 2\cos^2(x)\Big)\cos(3x)\cos(x) + \cos(6x) \Big] \end{align} $$ This allowed me to cancel the $\sin x$ in the numerator and denominator and compute the limit as $(1/7)$ by direct substitution, but as you can see this is not really a neat way of computing. Are there any other possible approaches?

Answer 1

Write

$$\frac{\sin x}{\sin 7x}=\frac{7x}{\sin 7x}\cdot\frac{\sin x}x\cdot\frac17$$

Answer 2

If you can use the fact that $\lim\limits_{x\to 0} \frac{\sin(x)}{x} = 1$, then here's a simple solution:

$$ \lim\limits_{x\to 0} \frac{\sin(x)}{\sin(7x)} = \frac{1}{7} \lim\limits_{x\to 0} \frac{\sin(x)}{x} \frac{7x}{\sin(7x)} = \frac{1}{7} $$

Answer 3

Though the solution using $\sin x/x$ is immediate, you can work this out with complex numbers.

Let $z=e^{ix}$, then

$$\lim_{x \to 0} \frac{\sin x}{\sin7x}=\lim_{z\to1}\frac{z-z^{-1}}{z^7-z^{-7}}=\lim_{z\to1}\frac1{z^6+z^4+z^2+1+z^{-2}+z^{-4}+z^{-6}}=\frac17.$$

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By the way, this tells you that

$$\frac{\sin 7x}{\sin x}=2(\cos6x+\cos4x+\cos2x)+1.$$

Answer 4

These answers are great, but I was reading a hint given on a completely different question: Find $\lim \limits_{x\to 0}{\sin{42x} \over \sin{6x}-\sin{7x}}$. I found it more intuitive that if I first write$$\frac{\sin(x)}{\sin(7x)} = \frac{\frac{\sin(x)}{7x}}{\frac{\sin(7x)}{7x}} = \frac{\frac{1}{7}\frac{\sin(x)}{x}}{\frac{\sin(7x)}{7x}},$$ I can then note that $\lim_{x\to0} \frac{1}{7}\frac{\sin(x)}{x} = 1/7$ and $\lim_{x\to0} \frac{\sin(7x)}{7x} = 1$. Of course, this still implies the form $$\frac{\frac{\sin(x)}{7x}}{\frac{\sin(7x)}{7x}} = \frac{\sin x}{7x}\cdot \frac{7x}{\sin(7x)} = \frac{7x}{\sin (7x)}\cdot\frac{\sin x}x\cdot\frac{1}{7}$$ but the intermediate step hint I got from the other answer is how I best understood it.

Answer 5

Your approach can be made neater using induction. We prove the general result.

Let $n$ be a positive integer. Then $\lim\limits _{x\to 0}\dfrac{\sin nx} {\sin x} =n$.

Clearly the result holds for $n=1$. And let's suppose that it holds for $n=m$ so that $\lim\limits _{x\to 0}\dfrac{\sin mx} {\sin x} =m$. Now we have $$\lim_{x\to 0} \frac {\sin (m+1)x}{\sin x} =\lim_{x\to 0} \frac{\sin mx\cos x+\cos mx\sin x} {\sin x} =\lim_{x \to 0} \frac{\sin mx} {\sin x}\cdot \cos x+\cos mx=m\cdot 1+1=m+1$$ Here we have used the limit $\lim\limits _{x\to 0}\cos x=1$. Also note that the fundamental result $\lim\limits_{x\to 0}\dfrac {\sin x} {x} =1$ has not been used here.

Thus the result holds for $n=m+1$ given the assumption that it holds for $n=m$. By induction the result holds for all positive integer values of $n$.

With a little more effort we can extend the result to all rational values of $n$ (and you may try to prove this). And now you can that the answer to your question is $1/7$.