Limit of $\sin(3x)/x$ as $x$ goes to zero I was asked to help a student with this limit as X goes to zero. 

$$\lim_{x \rightarrow 0}\frac{\sin \left(3x\right)}{x}$$

Note, I am able to solve it myself using L'Hopital's rule, just looking at a graph, or by the calculator method of sneaking up on the result by entering .1, .01, etc. 

But, the student told me the teacher wanted him to use the 'algebraic' method. I am a Math Aide (i.e. an in house tutor) and often the students will not have a text, only notes from class and worksheets. So I have no reference, and no opportunity to check with the teacher. Any assistance for what method the teacher wants? Keep in mind, the students haven't learned derivatives yet.
The limit
$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$
is an important and standard one, but proving it takes a bit of ingenuity. There is a beautiful answer here: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Assuming that limit has been established, we can apply it to this problem. Note that
$$\frac{\sin(3x)}{x} = 3\frac{\sin(3x)}{3x}$$
If we let $u = 3x$ then notice that $u \rightarrow 0$ if and only if $x \rightarrow 0$, so
$$\lim_{x \rightarrow 0}\frac{\sin(3x)}{x} = 3\lim_{x \rightarrow 0}\frac{\sin(3x)}{3x} = 3 \lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 3 \times 1 = 3$$
You could also write sin(3x) as sin(2x+x), which is equivalent to 
sin(2x)cos(x)+cos(2x) sin(x).
Then you could fill in the double angles:
2sin(x)cos(x)cos(x)+[1-2sin^2(x)]sin(x)
Simplify to
2sin(x)[1-sin^2(x)]+sin(x)-2sin^3(x)
Which is 
3sin(x) -4sin^3(x)
Now we have 
lim_x->0[3sin(x) -4sin^3(x)]/x]
3lim_x->0[sin(x)] - 4lim_x->0[sin(x)/x]•lim_x->0[sin^2(x)]]
3(1)-4(1)(0)=3
Or you can simply use the Sandwich Theorem, though that is not considered an algebraic solution per say.