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Lightning55 · Aug 23, 2009, 10:53 AM

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 )

galactus · Aug 23, 2009, 11:23 AM

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.

Lightning55 · Aug 23, 2009, 11:45 AM

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, $sin(200\pi x) = 0$

galactus · Aug 23, 2009, 12:52 PM

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 $\lim_{x\to 0}\frac{sin(ax)}{x}=a$