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I have some doubts on the theorem for substitutions of the variable in limits. There are two equivalent theorems for this.

The first one:

Suppose that $\lim_{x\to c} f(x)=l$ exists whith $c,l\in \mathbb{R} \cup \big\{+\infty,-\infty \big\}$, than let $g$ be a function defined in a neighbourhood $I(l)$ besides (at most) the point $l$, such that:

.If $l\in \mathbb{R}$, $g$ is continous in $l$

.If $l=\big\{+\infty,-\infty \big\}$, the limit $\lim_{y\to l} g(y)$ exists

Then $\lim_{x\to c} g(f(x))=\lim_{y\to l} g(y)$

The second one:

Suppose that $\lim_{x\to c} f(x)=l$ and $\lim_{y\to l} g(y)=m$ exist whith $c,l,m\in \mathbb{R} \cup \big\{+\infty,-\infty \big\}$ and that, if $l\in \mathbb{R}$, there exists a neighbourhood $I(c)$ in which $f(x) \neq l$ $ \forall x\in I(c)$ and $x\neq c$

Then again $\lim_{x\to c} g(f(x))=\lim_{y\to l} g(y)$

I don't understand in particular how the second theorem can be equivalent to the first one, and it is not clear to me how the condition for which exists a neighbourhood $I(c)$ in which $f(x) \neq l$ $ \forall x\in I(c)$ and $x\neq c$ it is related to existance of the limit.

Can anyone help me with this?

Thanks a lot for your help

Gianolepo
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  • I am unsure if the two statements are equivalent. As far as I know, if either assumption holds, then you can change variables in the limit. In the second statement, if $f$ is constant then nothing can be said about the limit of $g \circ f$. – Siminore Dec 26 '15 at 13:05
  • Thanks a lot for the answer! It's wrong to say that they are equivalent of course, I just wanted to say that the two theorems gives both the possibility to change variable. Anyway I can't understand why the fact that $f$ is constant is a problem, could you suggest me an example of that? – Gianolepo Dec 26 '15 at 13:40
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    Well, just define $g(y)=0$ for $y \neq 0$ and $g(0)=1$. Of course $\lim_{y \to 0} g(y)=0$. But if $f \equiv 0$, then $g \circ f(x) \to 1$ as $x \to 0$. – Siminore Dec 26 '15 at 17:51

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