Okay I saw this limit question in Thomas' Calculus 12th Edition:
$\lim \limits_{x \to 0} \frac{\tan3x}{\sin8x}$
The answer is $\frac{3}{8}$.
I was able to get the correct answer using this ridiculously lengthy and messed up rubbish...
... but I have a feeling there should a simpler, more concise and more elegant solution. Perhaps there is something I am missing? Maybe a trig identity? Or a theorem?
If you know about equivalents, then you can say that $\tan(y)$ is equivalent to $y$ in zero and $\sin(y)$ is equivalent to $y$ in $0$. This gives that the limit you're looking is the same as the limit, when $x$ tends to $0$, of $\frac{8x}{3x}$, which is of course $\frac{8}{3}$.
Here is an alternative proof if you don't know about equivalents:
\begin{align} \frac{\tan(3x)}{\sin(8x)} &= \frac{3}{8} \frac{\tan(3x)}{3x} \frac{8x}{\sin(8x)} \\ &= \frac{3}{8} \frac{\sin(3x)}{3x} \frac{1}{\cos(3x)} \frac{8x}{\sin(8x)}. \end{align}
These computations are done for $x$ close to $0$ but not equal to $0$ so that the denominators don't vanish. You certainly know that $\frac{\sin(y)}{y}$ tends to $1$ when $y$ tends to $0$. By composition and the fact that $\cos(0)$ is $1$, this shows that the three fractions tend to $1$, hence the result.