How do I flip the fraction (5x)/sin5x so that it is (sin5x)/5x?

I need it to be with the sin on the top to be able to use a certain theorem but i can't figure out what to multiply/divide by to get it flipped over without changing the value. I'm trying to find the limit of (1-cos5x)/5x * (5x/sin5x). I know there is the theorem that tells me the first bit of the function is 0, but I can't seem to get the other side to flip over so I can use the theorem that says sinx/x =1

Question

How do I flip the fraction (5x)/sin5x so that it is (sin5x)/5x?

I need it to be with the sin on the top to be able to use a certain theorem but i can't figure out what to multiply/divide by to get it flipped over without changing the value. I'm trying to find the limit of (1-cos5x)/5x * (5x/sin5x).  I know there is the theorem that tells me the first bit of the function is 0, but I can't seem to get the other side to flip over so I can use the theorem that says sinx/x =1

anonymous · Answer

$$\lim_{x \rightarrow 0} (\ \frac{ 1-\cos5x }{ 5x })(\frac{ 5x }{ \sin5x })$$
is my actual problem and I know that the left size goes to 0, and i should be able to say the right side goes to 1, but the equation is flipped upside down from what the theorem says : $$\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1$$

phi · Answer

you  can write  a/b  as   1/(b/a)

anonymous · Answer

oh ok thanks. that's simple!