Question

I'm struggling with this problem.

lim h->0 of (sin(t+h)-sin(t))/h)

Answer 1

looks like a first principles derivative problem

Answer 2

So (sin(t+0)-1)/0... But what do i do with the sin(t)??

Answer 3

This is pretty much the proof of f'(sinx) = cosx.
You have to use:
sin(t+h) = sintcosh+sinhcost

Answer 4

\[\lim_{h->0} \frac{sin(t+h)-sin(t)}{h}\]
\[\lim_{h->0} \frac{(sin(h)cos(t)+sin(t)cos(h))-sin(t)}{h}\]
\[\lim_{h->0} \frac{sin(h)cos(t)+sin(t)(cos(h)-1)}{h}\]
\[\lim_{h->0}\left( cos(t)\frac{sin(h)}{h}+sin(t)\frac{(cos(h)-1)}{h} \right)\]
\[cos(t)\frac{sin(0)}{0}+sin(t)\frac{(cos(0)-1)}{0}\]

Answer 5

\[\lim_{h->0}\frac{\sin(t+h) -\sin(t)}{h} \rightarrow \lim_{h->0}\frac{sintcosh+sinhcost - sint}{h}\]

Now you factorize sint. And split it...
\[\lim_{h->0} \frac{sint(\cosh-1)}{h} + \lim_{h->0}\frac{\sinh}{h} cost\]

Now you can pull outside the sint over there and multiply by "h", sinh/h is 1, cost you bring it outside the limit, and you get "cost"

Answer 6

they tend to apply the "squuze" thrm to this to suggest that those limits go to what they need to

Answer 7

*squeeze

Answer 8

ALFIE, "sinh/h is 1". This is wrong. Sinh/h is not 1.....???wtf

Answer 9

lim for h -> 0 of sinh/h is 1.

Answer 10

Proof is...:
sinx < x < tgx
sinx < x < sinx/cosx
divide by sinx..
1 < x/sinx < 1/cosx
cosx < sinx/x < 1.
For the limit of x -> 0... according to the squeeze theorem... sinx/x = 1...
Same stuff at your problem, just different name of variable.

Answer 11

Okay wow thanks.

I have trouble seeing what to do after here:

sint cosh+cost sinh-sint/h

On my test tomorrow, I'm afraid I wouldn't have seen the need to split and factor. Tricky...