proof of the differential of sin x

actually, yes, I was just re-reading it and wasn't happy with that part.

Remember the law of limits: $\lim \left[ f(h) g(h) \right] = \left[\lim f(h)\right] \left[\lim g(h) \right]$

so do you mean that if we re-write as a product



and take the limits of each in turn, the product of those limits is the limit of the product?

So,



and




the the limit of the product is zero?

$\displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0$

This is the problem line. The statement you need to make for your proof to work is:

$\displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx$

$\displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0$

This is the problem line. The statement you need to make for your proof to work is:

$\displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx$

You can then cancel the h on top by the h on the bottom. I know it's not a rigorous proof by any means, but it's what makes it work for C3/C4.

It's like:
$\frac {0.0001 cos{x}}{0.0001} = cos{x}$
$\frac {0.00000009999 cos{x}}{0.00000009999} = cos{x}$

$\displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = 0$
And
$\displaystyle\lim_{h\rightarrow 0}\;\sin h\cos x = h cosx$
Are equivalent on the limit..

Try for yourself: sin(0.0001) x cos(x) = 0.000099999 cos)x)

Nope. His line was fine, it's your line which doesn't make sense.

Sorry, but your whole post is nonsense. I don't think you really understand how limits work.