Chain rule for limits

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

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 [tex]\delta _1[/tex] for which the following is true for all [tex]x[/tex] in the domain:

[tex]0<|x-M|<\delta _1\Rightarrow |f(x)-L|<\epsilon[/tex]

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

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

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

[tex]0<|x-c|<\delta \Rightarrow |g(x)-M|<\delta _1[/tex]

Thus, by transitivity:

[tex]0<|x-c|<\delta \Rightarrow |f(g(x))-L|<\epsilon[/tex]

QED
 

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

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 [tex]\delta _1[/tex] for which the following is true for all [tex]x[/tex] in the domain:

[tex]0<|x-M|<\delta _1\Rightarrow |f(x)-L|<\epsilon[/tex]

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

[tex]0<|g(x)-M|<\delta _1\Rightarrow |f(g(x))-L|<\epsilon[/tex]
You meant [tex]0< |f(g(x))-M|[/tex] of course.

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

[tex]0<|x-c|<\delta \Rightarrow |g(x)-M|<\delta _1[/tex]

Thus, by transitivity:

[tex]0<|x-c|<\delta \Rightarrow |f(g(x))-L|<\epsilon[/tex]

QED
It would be better to distinguish between the various "&delta"s: write [tex]\delta_1, \delta_2[/tex], 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.
 
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[tex]M[/tex] is defined as the limit of [tex]g(x)[/tex] when [tex]x \to c[/tex]. [tex]L[/tex] is defined as the limit of [tex]f(x)[/tex] when [tex]x \to M[/tex]. Both are written in the first paragraph of the first post. Intuitively, I'm trying to say that as [tex]x[/tex] goes to [tex]c[/tex], [tex]g(x)[/tex] goes to [tex]M[/tex]. But, because as [tex]x[/tex] goes to [tex]M[/tex] in [tex]f(x)[/tex], [tex]f(x)[/tex] goes to [tex]L[/tex], I say that as [tex]x[/tex] goes to [tex]c[/tex], [tex]f(g(x))[/tex] goes to [tex]L[/tex].
In other words:
[tex]\lim_{x\to c} \, f(g(x))=\lim_{x\to \lim_{x\to c} \, g(x)} \, f(x)[/tex]
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 [tex]x[/tex] in which all [tex]x[/tex], when given as the parameter for [tex]f(g(x))[/tex], give outputs which are less distant than [tex]\epsilon[/tex] from [tex]L[/tex].
Just like in a proof of the sum rule you could finish with a line like [tex]\delta =\min \left\{\delta _1,\delta _2\right\}[/tex].
 

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