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?

Note that 
$$\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(5x)} = 
\lim_{x \rightarrow 0} \frac{3\cos(3x)}{5\cos(5x)} = \frac{3}{5}$$
by L'Hopital's rule. You can use this fact in approach 1 since $\cos(3x) \rightarrow 1$ and $\cos(5x) \rightarrow 1$.
Details:
$$\begin{align}
\lim_{x \rightarrow 0} \frac{\sin(3x) \cos(5x)}{\sin(5x)\cos(3x)}
&= \left(\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(5x)}\right)
\left(\lim_{x \rightarrow 0} \frac{\cos(5x)}{\cos(3x)}\right) \\
&= \left(\frac{3}{5}\right) \left(\frac{1}{1}\right) \\
&= \frac{3}{5} \\
\end{align}$$
