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 $$\delta =\min \left\{\delta _1,\delta _2\right\}$$.