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Is there a way to solve $\lim_{x \to 0}\frac{\tan(3x)}{\sin(8x)}$ without using the trig identity $\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-\tan^2(x)}$. I want to know because I had to look up this trig identity in order to solve this limit, but if there is a simpler way, then I would like to know it. I don't want to use L'Hospital's Rule because it hasn't been introduced at this point in the book.

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$$\lim\limits_{x \to 0}\frac{\tan(3x)}{\sin(8x)} = \lim\limits_{x \to 0}\frac{\sin(3x)}{\sin(8x)}\cdot\frac{1}{\cos{(3x)}} = \frac{3}{8}\lim\limits_{x \to 0}\frac{\sin(3x)}{3x}\cdot\frac{8x}{\sin(8x)}\cdot\frac{1}{\cos{(3x)}} = \boxed{\displaystyle\frac{3}{8}}$$

This uses $$\lim_{x\to 0} \frac{\sin(ax)}{ax} = 1$$

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HINT:

$$\frac{\tan3x}{\sin8x}=\frac38\frac{\sin3x}{3x}\frac1{\dfrac{\sin8x}{8x}}\frac1{\cos3x}$$

Use $\displaystyle\lim_{h\to0}\frac{\sin h}h=1$ [Observe that the angle of sine, denominator, the limit variable are same $(h)$]

What you have done if the problem was like $$\lim_{x\to0}\frac{\tan(\pi x)}{\sin(3.5x)}=?$$

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Some hints:

  1. $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$
  2. $\displaystyle\lim_{x\to0}\frac{\tan x}{x}=1$

Note that $$\frac{tan 3x}{\sin 8x}=\frac{\tan 3x}{3x}\cdot\frac{8x}{\sin 8x}\cdot\frac{3}{8}.$$

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This problem is simpler and does not require the use of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. We have $$\begin{aligned}L &= \lim_{x \to 0}\frac{\tan 3x}{\sin 8x}\\ &= \lim_{x \to 0}\frac{\tan x(3 - \tan^{2}x)}{1 - 3\tan^{2}x}\cdot\frac{1}{2\sin 4x\cos 4x}\\ &= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{4\sin 2x\cos 2x\cos 4x}\\ &= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{8\sin x\cos x\cos 2x\cos 4x}\\ &= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{1}{\cos x}\cdot\frac{1}{8\cos x\cos 2x\cos 4x}\\ &= \frac{3}{1}\cdot\frac{1}{1}\cdot\frac{1}{8\cdot 1\cdot 1\cdot 1} = \frac{3}{8}\end{aligned}$$ We have used the limits $$\lim_{x \to 0}\tan x = 0, \lim_{x \to 0}\cos x = 1$$ which are on slightly simpler level compared to $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$.

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