Let's break down the fraction into multiple factors:
[tex] \cfrac{\sin(ax)\cos(bx)}{\sin(cx)} = \sin(ax)\cdot\cos(bx)\cdot \cfrac{1}{\sin(cx)} [/tex]
Now we will manipulate the expression (multiply and divide by the same quantitues) in order to be able to use the known limit
[tex]\displaystyle \lim_{x\to 0} \cfrac{\sin(x)}{x} = 1 [/tex]
Here's the manipulated expression:
[tex] \sin(ax)\cdot \cfrac{ax}{ax} \cdot \cfrac{1}{\sin(cx)}\cdot\cfrac{cx}{cx}\cdot\cos(bx) [/tex]
Rewrite the expression as
[tex] \cfrac{\sin(ax)}{ax} \cdot \cfrac{cx}{\sin(cx)} \cdot \cfrac{ax}{cx} \cdot \cos(bx) [/tex]
The first two factors tend to 1, because that's the limit we mentioned before. The third factor is simply a/c, because the x's cancel out. Finally, we have
[tex] \displaystyle \lim_{x\to 0} \cos(bx) = \cos(0)=1 [/tex]
So, the final answer is
[tex] \displaystyle \lim_{x\to 0} \cfrac{\sin(ax)}{ax} \cdot \cfrac{cx}{\sin(cx)} \cdot \cfrac{ax}{cx} \cdot \cos(bx) = 1 \cdot 1 \cdot \cfrac{a}{c} \cdot 1 = \cfrac{a}{c} [/tex]
