Problem:
Assume that $\lim_{x\to0} \frac{\sin(x)}{x}=1$
Show that the function $f(x)=\sin(x)$ is continuous at 0.
Attempt:
My intuition is that since the $\lim_{x\to0} \frac{\sin(x)}{x}=1$, then $\lim_{x\to0} \sin(x)$ and $\lim_{x\to0} x$ are the same. So $\lim_{x\to0} \sin(x)=\lim_{x\to0} x=0$ and thus $\sin(x)$ is continuous at $0$.
I am not even sure if this is the correct way of thinking about the problem, and I don't know how to formulate a formal answer.
I would normally use the theorem: if $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$ , then $\lim_{x\to a}f(x)g(x)=l \cdot m$
However, in this particular case I cannot do this because $\lim_{x\to0} \frac{1}{x}$ does not exist.
I tried using: if $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$ , then $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{l}{m}$ but then I would be dividing by $0$ so it does not work either.
How else can I approach the problem? I know it is possible to just show $\sin(x)$ is continuous at 0 but how can the assumption be useful in this problem?
Your argument idea is fine. Use $f(x)=x$ and $g(x)=\frac{\sin x}{x}$. Then $$\lim_{x\to0}\sin x=\lim_{x\to0} f(x)g(x)=\lim_{x\to0} f(x)\lim_{x\to0} g(x)=0\cdot 1=0=\sin 0$$ as desired.
Another way to look at the probem is to note that $\frac{\sin x}{x}=\frac{\sin x-\sin 0}{x-0}$ so that the given limit tells us that $\sin'(0)$ exists. Differentiability implies continuity.