Thread: Differentiation via limits problem

Hi all, need some help please!

The question is as follows:

Suppose f(x) is differentiable at x = a, show that

lim as h->0 of (f(a+h)-f(a-h)) / 2h =f'(a)

I tried working from the difference quotient,

i.e: lim as h->0 of f(a+h) - f(a) / h, and i took this +h as the limit from the right side, and f(a-h) - f(a) / -h as limit from left side...is that along the right lines?

basically, you take the negative from that h and multiply the top part of the expression, and get -f(a-h) + f(a) / h for the left side, and since we know both are equal o f'(a), we can add them and say they, when added, are equal to 2f'(a), and the f(a) and -(fa) simply cancel out and you get:

2f'(a) = f(a+h) - f(a-h) / h

Then take the 2 across to get:

f'(a) = f(a+h) - f(a-h) / 2h

Does that make sense? Would i get any marks i wonder, is the logic 'followable'?...Please let me know! I'm sure there must be other ways too...

This is the usual way it is done.

Thanks for that, i managed to put in other steps to what you said and it seems right...I just need a little explanation about something, because although it is right (i had the math lecturer look at it) I don't actually understand why it is right, or where certain things come from (lecturer wasn't too clear on that...)

For example you can take:

lim h->0 (f(a+h)-f(a) / h),and you also take:

lim h->0 (f(a)-f(a-h) / h)

Where do these two come from? I mean the first one is just the quotient derivative as we're taught it, simply replacing x with a here though...but the second one?

How do you get the second one? Is it the limit as you approach from the left or something? (This is what i thought, but apprently that's wrong?...)

If someone can please explain what that second equation is and why it's equal to the first, and hence equal to f ' (a)?...

Thanks in advance once again!

Originally Posted by scorpio1

Thanks for that, i managed to put in other steps to what you said and it seems right...I just need a little explanation about something, because although it is right (i had the math lecturer look at it) I don't actually understand why it is right, or where certain things come from (lecturer wasn't too clear on that...)

For example you can take:

lim h->0 (f(a+h)-f(a) / h),and you also take:

lim h->0 (f(a)-f(a-h) / h)

Where do these two come from? I mean the first one is just the quotient derivative as we're taught it, simply replacing x with a here though...but the second one?

How do you get the second one? Is it the limit as you approach from the left or something? (This is what i thought, but apprently that's wrong?...)

If someone can please explain what that second equation is and why it's equal to the first, and hence equal to f ' (a)?...

Thanks in advance once again!

It's sort of an approach from the left. But it is easier to see as replacing "a" in your first expression with "a - h." It's really the same formula, just with a shift of variables.

-Dan

Hah...i just had that moment when you suddenly get something!

It is indeed the same as you say because it's simply a shift of variables.

I.e. lets say f(a+h) - f(a) will give you a slightly positive number...

then f(a) - f(a-h) should give you the same positive number since f(a) is obviously bigger than f(a-h), and by the same h as in the f(a+h) - f(a)!

Um, don't know if that made sense, but i'm just happy i get it!

Thanks for the help...oh and er, my logic is right there, right? O_o

Originally Posted by scorpio1

Hah...i just had that moment when you suddenly get something!

It is indeed the same as you say because it's simply a shift of variables.

I.e. lets say f(a+h) - f(a) will give you a slightly positive number...

then f(a) - f(a-h) should give you the same positive number since f(a) is obviously bigger than f(a-h), and by the same h as in the f(a+h) - f(a)!

Um, don't know if that made sense, but i'm just happy i get it!

Thanks for the help...oh and er, my logic is right there, right? O_o

More or less correct.

-Dan