Doubt regarding splitting of the limit $ \lim\limits_{x \to 0} \dfrac{\sin^2(3x)}{x^2} $ I have a simple question about the logic of the limits of trig functions. For example, let's say you have the following question:
$ \lim\limits_{x \to 0} \dfrac{\sin^2(3x)}{x^2} $
What's preventing me from doing this:
$ \lim\limits_{x \to 0} \sin(3x)\dfrac{\sin(3x)}{x^2} $
and then simplifying it to this: $ \lim\limits_{x \to 0} \sin(3x) \times  \lim\limits_{x \to 0} \dfrac{\sin(3x)}{x^2} $ and finally this: $0 \times \lim\limits_{x \to 0} \dfrac{\sin(3x)}{x^2}$
Thanks for the help.
 A: If $\lim_{x\to a} f(x) = L_1$ and $\lim_{x \to a} g(x) = L_2$ where $L_1, L_2$ are real numbers then you can conclude that $\lim_{x\to a } f(x) g(x) = L_1 L_2$. In this case, $\lim_{x\to 0} \frac{\sin 3x}{x^2}$ does not exist, so, your argument breaks down there.
A: The equality$$\lim_{x\to a}\left(f(x)g(x)\right)=\left(\lim_{x\to a}f(x)\right)\times\left(\lim_{x\to a}g(x)\right)\tag1$$holds when both limits $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist (and in some other cases too). In your case, the limit $\lim_{x\to0}\frac{\sin(3x)}{x^2}$ does not exist.
If you had $\lim_{x\to0}f(x)=r\in\Bbb R\setminus\{0\}$ and $\lim_{x\to0}g(x)=\pm\infty$ then $\lim_{x\to\infty}\bigl(f(x)g(x)\bigr)$ would be $\pm\infty$ if $r>0$ and it would be $\mp\infty$ is $r<0$. But, as I wrote above, the limit $\lim_{x\to0}\frac{\sin(3x)}{x^2}$ does not exist. It's not even $\infty$ or $-\infty$.
A: Maybe this gives another perspective:
$$0*\lim\limits_{x \to 0} \dfrac{\sin(3x)}{x^2}$$
Applying l'Hôpital's theorem:
$$0*\lim\limits_{x \to 0} \dfrac{3\cos(3x)}{2x}$$
Now, $$\lim\limits_{x \to 0} \dfrac{3\cos(3x)}{2x}$$ tends to infinity, so you basically get
$$0*\infty$$
which is an indeterminate form, reason why you cannot split like that. The splitting itself is not faulty but that doesn't lead you to anything practical.
