<s>[INST] How to find $\lim_{x\to 0}\frac{\tan 3x}{\tan 5x}$? I am asked to find the following limit:
$$ \lim_{x \to 0} \frac{\tan 3x}{\tan 5x}$$
My problem is in simplifying the function. I followed two different approaches to solve  the problem. But both seems incorrect.
Apprach 1)
Since $\tan \theta = \frac{sin \theta}{cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we have:
$$\frac{\tan 3x}{\tan 5x} = \frac{\sin 3x \times \cos 5x}{\sin 5x \times \cos 3x}$$
This approach does not work well, because I cannot simplify more. 
Approach 2)
Since $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \times \tan y)}$, we have:
$$\frac{\tan 3x}{\tan 5x} = \frac{\tan 3x}{\tan 3x + \tan 2x} = \frac{\tan 3x}{\frac{\tan 2 + \tan3}{1 - \tan 2 \times \tan 3}}$$
What am I doing wrong?
 [/INST] If the only "trigonometric limit law" you have available is $ \ \lim_{u \rightarrow 0} \ \frac{\sin \ u}{u} \ = \ 1 \ $ , you can take the expression apart as
$$  \lim_{x \to 0} \ \ \frac{\tan 3x}{\tan 5x} \ = \ \lim_{x \to 0} \ \ \frac{\sin 3x}{\cos 3x} \ \cdot \ \frac{\cos 5x}{\sin 5x} $$
$$ = \ \lim_{x \to 0} \ \ \sin 3x \ \cdot \ \frac{3x}{3x} \ \cdot \ \frac{1}{\cos 3x} \ \cdot \ {\cos 5x} \ \cdot \ \frac{5x}{5x} \ \cdot \ \frac{1}{\sin 5x} $$
$$ = \ \lim_{x \to 0} \  \frac{\sin 3x}{3x} \ \cdot \ \frac{1}{\cos 3x} \ \cdot \ \cos 5x \  \cdot \ \frac{5x}{\sin 5x} \ \cdot \ \frac{3x}{5x} $$
$$ = \ \lim_{x \to 0} \  \frac{\sin 3x}{3x} \ \cdot \ \lim_{x \to 0} \ \frac{1}{\cos 3x} \ \cdot \ \lim_{x \to 0} \ \cos 5x \  \cdot \ \lim_{x \to 0} \ \frac{5x}{\sin 5x} \ \cdot \ \frac{3}{5} $$
$$ = \ \lim_{3x \to 0} \  \frac{\sin 3x}{3x} \ \cdot \ \lim_{x \to 0} \ \frac{1}{\cos 3x} \ \cdot \ \lim_{x \to 0} \ \cos 5x \  \cdot \ \lim_{5x \to 0} \ \frac{1}{(\sin 5x) / 5x} \ \cdot \ \frac{3}{5} $$
$$ = \ 1 \ \cdot \ 1 \ \cdot \ 1 \  \cdot \ \frac{1}{\lim_{5x \to 0} \ [ (\sin 5x) / 5x]} \ \cdot \ \frac{3}{5} \ = \ \frac{1}{1} \ \cdot \ \frac{3}{5} \ = \ \frac{3}{5} \ \ . $$
We can generalize such arguments to produce limit laws such as  $ \ \lim_{x \rightarrow 0} \ \frac{\sin \ ax}{\sin \ bx} \ = \ \frac{a}{b} \ $ and $ \ \lim_{x \rightarrow 0} \ \frac{\tan \ ax}{\tan \ bx} \ = \ \frac{a}{b} \ $ .
(Reading over all the responses, this is basically the fully-detailed argument of what lab bhattacharjee shows, which is what you'd likely have to write out in an exam or homework problem early in  first-semester calculus.  It "builds character" and makes you grateful for L'Hopital...)
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