Why Are Assumptions Critical in the Limit of Composite Functions?

[sonofagun]

I need help with the following theorem:

Let I, J ⊆ℝ be open intervals, let x∈I, let g: I\{x}→ℝ and f: J→ℝ be functions with g[I\{x}]⊆J and Lim_z→xg(x)=L∈J. Assume that lim_y→L f(y) exists and that g[I\{x}]⊆J\{g(x)},or, in case g(x)∈g[I\{x}] that lim_y→L f(y)=f(L). Then f(g(x)) converges at x, and lim_z→x f(g(x))=lim_y→Lf(y).

I don't get why the assumptions are necessary. We're assuming that the limit of f exists when g is not defined at x, or, if g is defined at x, that the lim_y→L f (y)=f(L).

 

[Stephen Tashi]

I don't get why the assumptions are necessary.

What assumptions would you propose to use instead?

Do you mean "necessary" in the mathematical sense (as in "necessary and sufficient") or are you asking why certain assumptions are needed to write the proof the textbook expects?

 

[sonofagun]

What assumptions would you propose to use instead?

Do you mean "necessary" in the mathematical sense (as in "necessary and sufficient") or are you asking why certain assumptions are needed to write the proof the textbook expects?

I guess what I'm struggling to understand is why the existence of the limit of f(g(x)) depends on these assumptions. For example, if lim_y→L f(y) did not exist or if g[I\{x}] was not contained in J\{g(x)} would that imply that the limit of f(g(x)) doesn't exist? If so, why?

 

[Stephen Tashi]

For example, if lim_y→L f(y) did not exist or if g[I\{x}] was not contained in J\{g(x)} would that imply that the limit of f(g(x)) doesn't exist?

No, it wouldn't.

For example, define $f(x)$ by $f(x) = 0$ if $x < 0$ and $f(x) = 1$ if $x \ge 0$.
$Lim_{y \rightarrow 0} f(y)$ does not exist.

Define $g(x) = |x|$

$lim_{y \rightarrow 0} f(g(y)) = 1$

You haven't explained what you mean by "necessary".

There is a technical meaning for "necessary". The theorem that "A implies B" asserts that statement(s) A are "sufficient" for statement B to be true. To say that statement(s) A are "necessary" for B to be true asserts that "B implies A" .

There is a colloquial meaning for "necessary". By that meaning, we only assert that, having a certain method of proof in mind, the statements A are required to do that particular method of proof.

I don't know whether the "if..." part of the theorem you stated is "necessary" is the technical sense. The fact that theorem is stated in a text doesn't imply that the converse of the theorem is true.

( In the theorem, $lim_{z \rightarrow x} g(x) = L \in J$ should say "$g(z)$" instead of $g(x)$ .)

 

[sonofagun]

No, it wouldn't.

For example, define $f(x)$ by $f(x) = 0$ if $x < 0$ and $f(x) = 1$ if $x \ge 0$.
$Lim_{y \rightarrow 0} f(y)$ does not exist.

Define $g(x) = |x|$

$lim_{y \rightarrow 0} f(g(y)) = 1$

You haven't explained what you mean by "necessary".

There is a technical meaning for "necessary". The theorem that "A implies B" asserts that statement(s) A are "sufficient" for statement B to be true. To say that statement(s) A are "necessary" for B to be true asserts that "B implies A" .

There is a colloquial meaning for "necessary". By that meaning, we only assert that, having a certain method of proof in mind, the statements A are required to do that particular method of proof.

I don't know whether the "if..." part of the theorem you stated is "necessary" is the technical sense. The fact that theorem is stated in a text doesn't imply that the converse of the theorem is true.

( In the theorem, $lim_{z \rightarrow x} g(x) = L \in J$ should say "$g(z)$" instead of $g(x)$ .)

I was using the colloquial meaning of necessary.