Is My Chain Rule for Limits Proof Correct?

[Noxerus]

I would like to prove a chain rule for limits (from which the continuity of the composition of continuous functions will clearly follow): if \lim_{x\to c} \, g(x)=M and \lim_{x\to M} \, f(x)=L, then \lim_{x\to c} \, f(g(x))=L.

Can someone please tell me if the following proof is correct? I am a complete newbie to writing proofs, so I might have made several basic mistakes.


The second postulate means that there exists a \delta _1 for which the following is true for all x in the domain:

0<|x-M|<\delta _1\Rightarrow |f(x)-L|<\epsilon

By substituting x with g(x) we get the following which is true for all g(x) in the domain:

0<|g(x)-M|<\delta _1\Rightarrow |f(g(x))-L|<\epsilon

The first postulate means that there exists a \delta for which the following is true for all x in the domain:

0<|x-c|<\delta \Rightarrow |g(x)-M|<\delta _1

Thus, by transitivity:

0<|x-c|<\delta \Rightarrow |f(g(x))-L|<\epsilon

QED

 

[HallsofIvy]

I would like to prove a chain rule for limits (from which the continuity of the composition of continuous functions will clearly follow): if \lim_{x\to c} \, g(x)=M and \lim_{x\to M} \, f(x)=L, then \lim_{x\to c} \, f(g(x))=L.

Can someone please tell me if the following proof is correct? I am a complete newbie to writing proofs, so I might have made several basic mistakes.


The second postulate means that there exists a \delta _1 for which the following is true for all x in the domain:

0<|x-M|<\delta _1\Rightarrow |f(x)-L|<\epsilon

By substituting x with g(x) we get the following which is true for all g(x) in the domain:

0<|g(x)-M|<\delta _1\Rightarrow |f(g(x))-L|<\epsilon

You meant 0< |f(g(x))-M| of course.

The first postulate means that there exists a \delta for which the following is true for all x in the domain:

0<|x-c|<\delta \Rightarrow |g(x)-M|<\delta _1

Thus, by transitivity:

0<|x-c|<\delta \Rightarrow |f(g(x))-L|<\epsilon

QED

It would be better to distinguish between the various "&delta"s: write \delta_1, \delta_2, etc.
Also, it's not clear what "M" is. You started by writing |x-M| which should have been, from your statement of the theorem, c.

 

[Noxerus]

M is defined as the limit of g(x) when x \to c. L is defined as the limit of f(x) when x \to M. Both are written in the first paragraph of the first post. Intuitively, I'm trying to say that as x goes to c, g(x) goes to M. But, because as x goes to M in f(x), f(x) goes to L, I say that as x goes to c, f(g(x)) goes to L.
In other words:
\lim_{x\to c} \, f(g(x))=\lim_{x\to \lim_{x\to c} \, g(x)} \, f(x)
Thus I believe that my original statements are correct as written.

As for the deltas, I didn't give a subscript to the second delta because it is the "final" delta, i.e. the distance around x in which all x, when given as the parameter for f(g(x)), give outputs which are less distant than \epsilon from L.
Just like in a proof of the sum rule you could finish with a line like \delta =\min \left\{\delta _1,\delta _2\right\}.