I'm trying to find the solution for the following limit without using L'Hopitals rule.
The indeterminate form of $\frac{0}{0}$ is obtained but both the conjugate and or squeeze theorem can't be applied here (I think). I know that the solution is supposed to be 3 but I can't see how to reach it.
$\lim \limits_{x \to 0} \frac{sin3x}{x}$
$$\lim \limits_{x \to 0} \frac{\sin3x}{x}=\lim \limits_{x \to 0} 3\frac{\sin3x}{3x}$$