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On limit of function and proof of chain rule

  1. Sep 6, 2014 #1
    Definition of 'Limit of function (f) at x=p'

    Let E be domain of f and p be a limit point of E. Let Y be the range of f.

    If there exists q∈E such that for all ε>0 there exists δ>0 such that for all t∈E for which d(t,p)<δ implies d(f(t),q)<ε. Then we say that f(t)->q as t->p.

    1) Suppose f is continuous on [a,b] and g is defined on an interval I which contains the range of f.
    Since f is continuous on [a,b] for any x in [a,b], as t->x, f(t)->f(x).
    Let s be an element of I. Then can we say that as t->x. s->f(x)?

    2) Suppose f is continuous on [a.b], f'(x) exists at some point x in [a,b]. g is defined on an
    interval I which contains the range of f and g is differentiable at the point f(x).
    This is what i want to prove: g'(f(x))=g'(f(x))f'(x)

    This is what i have tried:
    [g(f(t))-g(f(x))]/[t-x]= [g(f(t))-g(f(x))]/[f(t)-f(x)] * [f(t)-f(x)]/[t-x]
    Letting t->x, we see that f(t)->f(x). Thus, g'(f(x))= g'(f(x)) * f'(x)
    Is something wrong with this proof?
     
  2. jcsd
  3. Sep 6, 2014 #2

    Fredrik

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    It doesn't make sense to write s→f(x). For example, if I=[0,4], s=3 and f(x)=5, this would say that 3→5.

    This is a good way to guess what the result will be, but it's not a proof. To say that ##f(t)\to f(x)## as ##t\to x## is to say that f has a limit at x, and that limit is f(x). This is a statement about f, so you can't immediately conclude that it ensures that ##\frac{g\circ f-g(f(x))}{f-f(x)}## has a limit at x.
     
  4. Sep 7, 2014 #3
    In here, I must contain the range of f, thus f(x) cannot be 5. And also, s does not refer to specific number in I, like t in t->x
     
  5. Sep 7, 2014 #4

    Fredrik

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    You're right. My mistake. So let me correct my example: If I=[0,6], s=3 and f(x)=5, this would say that 3→5

    You said "Let s be an element of I". That comment makes it a specific number in I (assuming that I is a set of numbers).
     
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