\(\displaystyle {\lim }\limits_{x \to 0} \frac{{\sin (3x)}}{x}\)
let z = 3x
(x = z/3) (**don't forget to change the limit, here it is easy,
as x approaches 0, z approaches 0 also.)
\(\displaystyle \begin{array}{l}
{\lim }\limits_{z \to 0} \frac{{\sin (z)}}{{\frac{z}{3}}} \\
{\lim }\limits_{z \to 0} 3\frac{{\sin (z)}}{z} \\
3\left( {{\lim }\limits_{z \to 0} \frac{{\sin (z)}}{z}} \right) \\
3\left( 1 \right) \\
3 \\
\end{array}\)
Show us what you have done so far with the implicit functions. This isn't a homework doing forum, this is a concept understanding forum.
Please post your progress on those questions, we aren't here to laugh at silly errors. Pointing out ones mistakes is an effective learning tool.