Well, so I have a question that says f(x)=sin(200\pi x)
Then it gives the instantaneous rate of change as \lim_{x\to\0}=\frac{f(x)-f(0)}{x-0}

After plugging in the equations, I don't know how to rid the denominator of x.

\lim_{x\to\0}\frac{sin(200\pi x)}{x}

(I'm finally getting this math code thing :D )

Make use of the famous limit:

\lim_{x\to 0}\frac{sin(x)}{x}=1

What is your limit then?

The above limit is usually proven using the Squeeze Theorem. It may be a little daunting. That is why it is often just used as an identity because it pops up so much.

I'm still not getting everything.

sin(200\pi) = 0

But since

sin(200\pi x) = sin(x)

(Is that right?)
Then

\lim_{x\to\0}\frac{sin(200\pi x)}{x} = \lim_{x\to\0}\frac{sin(x)}{x} = 1

Right?

But on my calculator, sin(200\pi x) = 0

Per the limit I posted previously:

\lim_{x\to 0}\frac{sin(200{\pi}x)}{x}=200{\pi}

Look at the limit \lim_{x\to 0}\frac{sin(1\cdot x)}{x}=1

See the similarity? See why the limit of the first is 200{\pi}?

It's rather obvious using the last limit as a identity.

If we had \lim_{x\to 0}\frac{sin(ax)}{x}=a