<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] Using L'Hopital's:
$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \lim \limits_{x \to 0} \frac{3 \sec^2 3x}{5 \sec^2 5x} = \frac{3}{5}$$
Without L'Hopital's:
$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \frac{3}{5} \lim \limits_{x \to 0} \frac{\cos 5x}{\cos 3x} \cdot \frac{\sin 3x}{3x} \cdot \frac{5x}{\sin 5x} = \frac{3}{5}$$
This uses the fact that $\lim \limits_{u \to 0} \frac{\sin u}{u} = 1$. This approach derives the identities $\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac ab$ and $\lim_{x \to 0} \frac{\tan ax}{\tan bx} = \frac ab$.
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