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find a counterexample of this statement of expression about the limit of composition function

ianchenmu

Member
Feb 3, 2013
74
Suppose that $U$ is open in $\mathbb{R}^{m}$, that $L\in U$ and that $h:U\setminus \left \{ L \right \}\rightarrow \mathbb{R}^{p}$ for some $p\in N$. If $L=\lim_{x\rightarrow a}g(x)$ and $M=\lim_{y\rightarrow L}h(y)$. Then
$\lim_{x\rightarrow a}(h\circ g)(x)=M$.




(Someone told me that this statement is false and should replace "$U\setminus \left \{ L \right \}$" with "$U$", and “$M=\lim_{y\rightarrow L}h(y)$. Then”
with “$h$ is continuous at $L$, then”.)


So can you give me a counterexample of the original statement? Thanks.
 

jakncoke

Active member
Jan 11, 2013
68
What is the domain and range of g(x)?
 

ianchenmu

Member
Feb 3, 2013
74
What is the domain and range of g(x)?
Let $a\in \mathbb{R}^{n}$, let $V$ be an open set which contains $a$, and suppose that $g: \ V\setminus a\rightarrow \mathbb{R}^{m}$.
 
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jakncoke

Active member
Jan 11, 2013
68
Let $a\in \mathbb{R}^{n}$, let $V$ be an open set which contains $a$, and suppose that $f: \ V\setminus a\rightarrow \mathbb{R}^{m}$.
The domain of h is U \ L, so can i assume that $\mathbb{R}^{m}$ was a typo and $f: \ V\setminus a\rightarrow$ U \ L ?
 

ianchenmu

Member
Feb 3, 2013
74
The domain of h is U \ L, so can i assume that $\mathbb{R}^{m}$ was a typo and $f: \ V\setminus a\rightarrow$ U \ L ?

I'm sorry, it's $g$.

Let $a\in \mathbb{R}^{n}$, let $V$ be an open set which contains $a$, and suppose that $g: \ V\setminus a\rightarrow \mathbb{R}^{m}$.
 

jakncoke

Active member
Jan 11, 2013
68
First of all the continuity is a must, for if

$
f(x) = \begin{array}{cc} \{ & \begin{array}{cc} -1 & x < 0 \\ 1 & x \geq 0 \end{array} \end{array} $

then $lim_{n \to \infty} \frac{1}{n} \to 1$ and $lim_{n \to \infty} - \frac{1}{n} \to -1$ ($n \in \mathbb{N}$)
f(x) is not continous at x = 0.

so if you evaluate f(g(x)), $g(x) \in dom(F) $, then if $lim_{x \to a} g(x) \to 0$, then
$lim_{x \to a} f(g(x))$ could be either -1 or 1 (Depending on from which point (left or right) g(x) is heading towards a).

Which brings me to my next point, to verify continuity at a point $a$, you have to be able to evaluate it, namely it should be in the domain of f. if you remove it from the domain of f, like when you said $U-\{L\}$, then f(g(x)) $lim_{x \to a} f(g(x))$ g(x) is heading towards an undefined point, since $g(x) \to L$, how could you check continuity at an undefined point? Namely the definition of continuity is (f is continous at a point a in its domain) if $lim_{x \to a} f(x) = f(a)$, where x can approach from any direction. if f(L) is not defined then how can we check continuity?

Your statement holds true if modified as follows
Let U be an open subset of $\mathbb{R}^{p}$, and g: U $\to \mathbb{R}^{m}$ and g is continous at point $L \in U$
and let A be an open subet of $\mathbb{R}^{w}$, and h: A - {a} $ \to U$.
and assume $lim_{x \to a} h(x) = L$, and $lim_{y \to L} g(y) = M$, then it certainly holds that $lim_{x \to a} g(h(x)) = M$.

I also probably need to add the fact that a is not an isolated point of open set A. (can you tell me why?)
 
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