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. 
 A: 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
$$\lim_{x \rightarrow 0}\frac{\sin(3x)}{x} = 3\lim_{x \rightarrow 0}\frac{\sin(3x)}{3x} = 3 \lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 3 \times 1 = 3$$
A: From what I understand, the "algebraic method" refers to "algebraically simplifying the function before trying to evaluate its limit" (as seen here on page 4).
To algebraically simplify $\dfrac {\sin 3x}x$, write $\sin x$ in summation form:
$$\sin x=\sum_{i=0}^\infty (-1)^i\frac{x^{2i+1}}{(2i+1)!}\\
\frac{\sin 3x}x=\frac 1x\sum_{i=0}^\infty (-1)^i\frac{(3x)^{2i+1}}{(2i+1)!}=\sum_{i=0}^\infty (-1)^i\frac{3^{2i+1}x^{2i}}{(2i+1)!}=3-\frac {3^3x^2}{6}+\frac{3^5x^4}{120}-\dots$$
Then we have $\lim_{x\to 0}\dfrac{\sin 3x}x=3.$
A: I can only speculate, but I suspect that in this class the limit of $\sin(x)/x$ is taken as a known thing, and the "algebraic method" the student is supposed to use involves manipulating the given expression so that it can be expressed in terms of that known limit.
A: Just a note, since I don't seem to have the reputation to comment and was originally led astray by other comments on the original question, l'Hopital's rule CAN be applied (if it wasn't explicitly specified not to use it in the question), and in practice would have been the most practical approach to solving this or similar limits.
There is no requirement that the limit-definition of the derivative be defined without encountering the same limit, as @Emily seems to suggest.  I understand that there is a circular logic in trying to show
$$\left.\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)\right|_{x=0}=\lim_{x\to0}\frac{\sin(x)-sin(0)}{x-0}=\lim_{x\to0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)}{1}.$$
But as long as the exercise isn't to try to derive this derivative using limits alone, it's a perfectly valid argument.  The only requirements to l'Hopital's rule for solving $\lim_{x\to c}f(x)/g(x)$ are that $f$ and $g$ be differentiable on a small open interval $I$ around $c$, that $g^{\prime}\neq 0$ on that interval, and that $\lim_{x\to c}f^\prime(x)/g^\prime(x)$ exist.  All of these conditions are satisfied in the question.
In fact, l'Hopital's rule is very often used to show that $\lim_{x\to0}\mathrm{sinc}(x)=\frac{\sin(\pi x)}{\pi x}=1$.
A: 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.
