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Suppose we have a function $ f: X \to Y $ such that $ d_1 (f (x), y_0) <A_1 $ whenever $ d_0(x, x_0) <A_0 $ and so such a function $ g:Y \to W $ such that $ d_2 (g (y), w_0) <A_3 $ if $ d_1 (y, y_0) <A_4 $. $d_0,d_1,d_2$ are the metrics. It´s not true that the limit of $h$ can be calculated simply as the limit of g if and tends to $ y_0 $ (only this is not necessarily equal to $f$ and thus compose want to know what is wrong with the show) if poorly explain my doubt, see the example I put where it is shown

The question is what is wrong in the next proof, that try to demonstrate this false property?

False Proof:

Taking $ A_3 = A_2 $ exists $ A_0 $ such that: $ d_2 (g (y), w_0) <A_2 $ whenever $ d_1 (y, y_0) <A_1 $ but using the fact that $ d_1 (f (x), y_0) <A_1 $ whenever $ d_0(x, x_0) <A_0 $ we have that those $f (x)$ exist in the ball of $ d_1 (y, y_0) <A_0 $ therefore evaluating $g$ in those points will also be at a distance of less than $ A_2 $ of the point $ w_0 $ Therefore $ h (x) \to w_0 $ as $ x \to x_0 $ But this proof is bad! but i dont know why! counterexample

$ f(x) = 0 $ for x $ \in [0,1] $

$ g(0) = 0 $

$ g(x) =1 $ for $x \in (0,1] $

$ h(x) = g(f (x)) $ clearly $g(x) \to 1$ as $x \to 0$ but $g(f(x)) \to 0$ as $x \to 0$.

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If I understand your question correctly, your hypothesis is really this:

For every positive real number $A_1$ there is a positive real number $A_0$ such that $d_1(f(x),y_0) < A_1$ whenever $d_0(x,x_0) < A_0$, and for every positive real number $A_3$ there is a positive real number $A_4$ such that $d_2(g(y),w_0) < A_1$ whenever $d_1(x,x_0) < A_4$. In other words, $\lim\limits_{x \to x_0} f(x) = y_0$, and $\lim\limits_{y \to y_0} g(y) = w_0$.

And your desired conclusion is that $\lim\limits_{x \to x_0}h(x) = w_0$, where $h = g \circ f$.

If this is what you intend, your proof can be fixed up quite easily, and your example isn't a counterexample. In your example all three spaces are $[0,1]$ and all three metrics are simply $d(x,y) = |x - y|$, so I'll use the simpler notation. Take $w_0 = 0$ and $A_2 = 1/2$; there is no possible value of $A_1$ such that $|g(y)| < 1/2$ whenever $|y| < A_1$, since $g(y) = 1$ whenever $0 < y < A_1$. Thus, this $g$ doesn't satisfy the hypothesis of the theorem.

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I think your question is obscured by the explicit mention of various metrics. The actual question is the following: Confronted with a composite limit $$\lim_{x\to \xi} \ g\bigl(f(x)\bigr)\ ,$$ is it allowed to replace the "inner function" $f(x)$ by a new variable $y$ and the limit point $\xi$ by the limit $\eta:=\lim_{x\to\xi}\ f(x)$ of the inner function? Then you could simply compute the limit $$\lim_{y\to \eta} \ g(y)$$ instead. Your example shows that this can go wrong. Now there are two situations where the described procedure is o.k., and in a large part of the occurring cases (e.g., when $\eta=\pm \infty$) one of them is indeed realized. They are:

(a) The function $f$ does not assume the value $\eta$.

(b) The function $g$ is continuous at $\eta$.

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