Also, I can't figure out this problem on my homework, so I'm wondering if one of you can show me how to do it.
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Find limit f(x + h) - f(x) / h
for f(x) = 1/ (x-1)
I mean, I know you plug in it in for x.
So, I got to something like
f(1/(x-1) + h) - f(1/(x-1))
At this point, I multiplied h/1 by x-1 to get a common denominator. Ended up with
1+hx-h / x-1
And then, subtracted 1/(x-1) from that and ended with
[hx-h / x-1] / h
And then, I tried something here that yielded the wrong answer. (multiplied top and bottom by x-1) Soooo, I figure I must have done something wrong.
The right answer is supposed to be
-1 / (x-1)^2
Do you know the shortcut? You'll be very happy when you learn it, if you don't. =]
Um, but assuming h is just a variable, if you make it the change in x, then that's in the form of the formal definition of a derivative:
f'(x)=lim as change in x-->0 of (f(x + change in x) - f(x))/change in x
or f'(x)=lim as h-->0 of (f(x +h) - f(x))/h
To solve it, I'd use the two points: (x, f(x)) and (x+h, f(x+h)) (where h is the change in x)
For the first point, you look at your function f(x) = 1/ (x-1).
You know x is just x, and f(x) is just that function. So the first point is (x, 1/(x-1))
The second point is basically what the same point would be as the first one, but for the function f(x+h) instead of just f(x). So the second point will be (x+h,f(x+h)). To find f(x+h), you simply plug (x+h) into your function where x would be. So it would be: 1/((x+h)-1). The second point is then (x+h, 1/((x+h)-1)). Written out, this looks a lot less complicated than when it's typed.
So using these two points, you can fill in part of the function you're taking the limit of. Replacing f(x+h) with 1/((x+h)-1) from above, and f(x) with your initial function:
f'(x)=lim as h-->0 of (f(x +h) - f(x))/h
changes to
or f'(x)=lim as h-->0 of (1/((x+h)-1) - 1/ (x-1))/h
If you then simplify this you get:
f'(x)=lim as h-->0 of ((x-1)-(x+h-1))/((x+h-1)(x-1)(h))
This goes to:
f'(x)=lim as h-->0 of (x-1-x-h+1)/((x+h-1)(x-1)(h))
You'll notice that on top the x's and the 1's cancel out, leaving you with a negative h on top:
f'(x)=lim as h-->0 of -h/((x+h-1)(x-1)(h))
Then you'll notice the h on top and h on bottom cancel:
f'(x)=lim as h-->0 of -1/((x+h-1)(x-1))
At this point you plug in 0 (what h is going to in this limit), and solve. You can only plug in zero now because before now if you had plugged in zero for h, it would have given you an undefined value because you divide by h. Now that you've canceled the h term and the only h in it is a part of a term and not a term by itself, you can plug in 0 for h:
f'(x)=lim as h-->0 of -1/((x+0-1)(x-1))
f'(x)=lim as h-->0 of -1/((x-1)(x-1))
f'(x)=lim as h-->0 of -1/(x-1)^2
Edit: So what you were doing wrong was you were plugging in the function for x, when x is just x and the function is f(x).
Oh, and also, f'(x) is notation for derivative.
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