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

#### ianchenmu

##### Member
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
What is the domain and range of g(x)?

#### ianchenmu

##### Member
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
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
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
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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