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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?

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# On limit of function and proof of chain rule

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