Two limits with unknown functions $f,g$ If $$\lim_{x\to 0}{\frac{f(x)-1}{\sin(x)}}=2$$ and $$\lim_{x\to 0}{\frac{g(x)-1}{\sin(x)}}=3$$ then find $$\lim_{x\to 0}{\frac{f(x)\cdot{g(x)}-1}{\sin(x)}}$$ and $$\lim_{x\to 0}{\frac{x\cdot{f(x)}-g(x)\cdot{\sin(x)}-x+\sin(x)}{\sin^2(x)}}$$
My first question is if $\lim\limits_{x\to 0}{\frac{f(x)\cdot{g(x)}-1}{\sin(x)}}=5$ and if, in order to find the $\lim\limits_{x\to 0}{\frac{x\cdot{f(x)}-g(x)\cdot{\sin(x)}-x+\sin(x)}{\sin^2(x)}}$, I must factor with $x$ and $\sin(x)$ respectively and then  divide with $\sin(x)$
 A: $$\lim_{x\to 0}{[\frac{f(x)\cdot{g(x)}-1}{\sin(x)}]}=\\\lim_{x\to 0}{[\frac{f(x)\cdot{g(x)}-1-g(x)+g(x)}{\sin(x)}]}=\\\lim_{x\to 0}{[\frac{(f(x)-1){g(x)}+g(x)-1}{\sin(x)}]}$$divide into cases
$$=\\\lim_{x\to 0}{[\frac{(f(x)-1){g(x)}}{\sin(x)}]}+\lim_{x\to 0}{[\frac{g(x)-1}{\sin(x)}]}=\\\underbrace{\lim_{x\to 0}{g(x)}}_{w.r.t \space  \lim_{x\to 0}{\frac{g(x)-1}{\sin(x)}}=3 \implies g(0) \to 1}\lim_{x\to 0}{[\frac{(f(x)-1)}{\sin(x)}]}+\lim_{x\to 0}{[\frac{g(x)-1}{\sin(x)}]}=1.2+3=5$$ for the second one 
$$\lim_{x\to 0}{\frac{x\cdot{f(x)}-g(x)\cdot{\sin(x)}-x+\sin(x)}{\sin^2(x)}}=\\\lim_{x\to 0}{\frac{x({f(x)}-1)-(g(x)-1){\sin(x)}}{\sin^2(x)}}=\\
\lim_{x\to 0}{\frac{x({f(x)}-1)}{\sin^2(x)}}-\lim_{x\to 0}{\frac{(g(x)-1){\sin(x)}}{\sin^2(x)}}=\\\text{simplify   and you know },\lim_{x\to 0}\frac{\sin x}{x}=1\\
\lim_{x\to 0}{\frac{({f(x)}-1)}{\sin(x)}\cdot\frac{x}{\sin x}}-\lim_{x\to 0}{\frac{(g(x)-1)}{\sin(x)}}=\\2.1-3=-1$$
A: Observe
$$f(x)g(x)-1 = (f(x)-1)(g(x)-1) +(f(x)-1)+ (g(x)-1).$$
Dividing by $\sin x$ gives
$$\frac{f(x)g(x)-1}{\sin x} = \sin x \cdot\frac{(f(x)-1)}{\sin x}\cdot \frac{(g(x)-1)}{\sin x} +\frac{(f(x)-1)+ (g(x)-1)}{\sin x}.$$
The limit of the left side is thus $0\cdot 2\cdot 3 + 2 + 3 = 5.$ 
A: Redefine $f(0)=g(0)=1$. (The values of $f(0)$ and $g(0)$ are irrelevant for the limits, so you can redefine the values without changing the limit. This redefinition assures that $f,g$ are continuous at zero.) 
Noting that $\lim_{x\to 0}\frac{\sin x}{x}=1$, we see that the above means that $f'(0)=2$ and $g'(0)=3$. This is because:
$$\frac{f(x)-f(0)}{x}=\frac{f(x)-f(1)}{\sin x}\cdot\frac{\sin x}{x}$$
The limit of the left side is by definition $f'(0)$, the limit of the right is $2\cdot 1$.
Now:
$$\frac{f(x)g(x)-1}{\sin x}=\frac{f(x)g(x)-f(0)g(0)}{x}\cdot \frac{x}{\sin x}
$$
The limit requested is the equal to $$(fg)'(0)=\lim_{x\to 0}\frac{f(x)g(x)-f(0)g(0)}{x}$$
But by the product rule of derivatives, this is $f(0)g'(0)+f'(0)g(0)=1\cdot 3+2\cdot 1=5$.
