calculus - Find the limit as x approaches 0

$$\lim_{x\to 0} \frac{\sin(3x)}{x+\tan(4x)}$$

I am having trouble with this one.

I know that $\lim_{x\to 0}\frac{\sin(x)}{x} = 1$
and same for the inverse

also $\lim_{x\to 0}\frac{\tan(x)}{x} = 1$

I am able to simplify $\sin(x)$ easily by multiplying by $\frac{3x}{3x}$

I am lost as to what to do next however. I seems like a simple problem (and it probably is) and I also know that the limit of the whole function as $x$ approaches $0$ will be $3/5$.

I am also not allowed to use L'Hospital's rule.

Thanks!

Answer

Write

$$\frac{\sin(3x)}{x+\tan(4x)} = \frac{\sin(3x)/x}{1+\tan(4x)/x} = \frac{3\cdot \sin(3x)/(3x)}{1+4\cdot \tan(4x)/(4x)}$$

See what to do now?