<p>For a differentiable function f, let f* be the function defined by</p>
<p>f*= lim [ f(x+h) - f(x - h) ] / [ h ]
h->0 </p>
<p>(different from the definition of a derivative btw, it's not a typo)</p>
<p>determine f*(x) for f(x) = sin (x)</p>
<p>every time i try this problem I end up at [ 2 sin(h) cos (x) ] / h which would still give me 0/0 if I plug in...can anyone help?</p>
<p>well, actually it is the definition of the derivative as well.</p>
<p>So cos(x)</p>
<p>what?</p>
<p>isn't the definition of the derivative f ' (x) = f (x + h) - f(x) // h?</p>
<p>Yes, but this an alternate definition. This definition also makes a lot of sense when you look at it graphically.</p>
<p>i'm not arguing with you but if i take the derivative of f(x) = x using the formal definition I get 1, when I find the limit of f(x) = x (the limit in my original post) i get 2...did i do something wrong?</p>
<p>Acually it's not an alternate definition, unless you copied it wrong.</p>
<p>The alternate definition is lim h->0 f(x+h)-f(x-h)/2h . The two is needed to "average out" the numerator, so it's a lot like the difference quotient in its philosophy.</p>
<p>So technically what you are looking for, f*, is double f'. So the answer is 2cosx.</p>
<p>You also could have gotten this through the identity sin(a±b)=sinacosb±cosasinb and worked out the limit. That comes out to the same answer and is a little more rewarding (in a very dorky way haha).</p>
<p>EDIT: To reply to your previous post, you did nothing wrong. That was a really smart thing to do, actually. But it just serves as another example that your expression is the double of the derivative.</p>
<p>Oh haha, sorry for sounding hostile.</p>
<p>Actually it might be the form [f(x+h) - f(x-h)]/2h is the derivative, so your form might be 2f'(x)</p>
<p>so 2cosx?</p>
<p>oops, chris beat me to it</p>
<p>oh ok thanks, but has anyone tried to do it out by hand? i can't seem to do it using the sum and difference formulas...i keep getting stuck at 2sinhcosx / h...i guess i'll try again, thanks a lot</p>
<p>jfs4691, that last expression you have is right. </p>
<p>lim (2cosxsinh)/h =
h->0 </p>
<p>2cosx * lim sin(h)/h<br>
h->0 </p>
<p>since you can separate terms from the limit (2cosx does not involve the variable h, so its ok to take it out). The second part of the bottom limit, involving sinh/h, is a common limit that equals one. Therefore you have 2cosx*1, which is just 2cosx.</p>
<p>oh ok thanks a lot, i should have noticed that.</p>
<p>No problem.</p>