On limit of function and proof of chain rule

[jwqwerty]

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?

 

[Fredrik]

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)?

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.

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?

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.

 

[jwqwerty]

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.

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

 

[Fredrik]

In here, I must contain the range of f, thus f(x) cannot be 5.

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

And also, s does not refer to specific number in I, like t in t->x

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).

 

[blue_raver22]

1) Yes, we can say that as t->x, s->f(x). This is because as t approaches x, the value of f(t) approaches the value of f(x). Therefore, as t gets closer and closer to x, the value of s, which is dependent on f(t), also gets closer and closer to the value of f(x). This is a direct consequence of the definition of a limit of a function.

2) Your proof is correct. The use of the quotient rule and the fact that f(t)->f(x) as t->x is a valid approach to proving the chain rule. However, it may be helpful to provide a more detailed explanation of your steps and why they lead to the desired result. Additionally, it may be useful to mention the properties of continuity and differentiability in your proof to make the reasoning more explicit. Overall, your proof is valid and demonstrates a good understanding of the chain rule.