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A continuous real-valued function $f: A \to B$, such that $A, B \subseteq \mathbb{R}$ where $f(0)=1$ and $\displaystyle \lim_{x \to \infty} f(x)=-1$ must have a zero within $(0,\infty)$

However, if an asymptote existed at some $x = \alpha$, where $\alpha \notin A$ such that $f(x) \to \infty$ as $x \to \alpha^{+}$ and $f(x) \to -\infty$ as $x \to \alpha^{-}$, and in the interval $(\alpha, \infty)$, the function approaches the limit $f(x) = -1$ as $x \to \infty$, isn't there a possibility that a zero does not exist within $(0,\infty)$, thus disproving the statement?

EDIT: Also, what if $\alpha \notin A$ such that $f(x) \to \infty$ as $x \to \alpha^{-}$ and $f(x) \to -\infty$ as $x \to \alpha^{+}$? Sorry for the edit again.

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You need to tell us what $A$ and $B$ are, because if $A$ is not connected, the theorem is false as you noticed. If $A$ is connected (or at least contains the whole of $(0,\infty)$) then the result is true, can you prove it? –  Najib Idrissi Mar 17 '12 at 8:48
    
You shouldn't change the question after you get answers. Eventually write that you want something else at the end of the answer, but do not make the ones who gave a part of their time to give some nice answers, look like they answered another question. –  Beni Bogosel Mar 17 '12 at 10:56
    
And you could motivate why do you really want that the function doesn't have a zero? –  Beni Bogosel Mar 17 '12 at 10:58
    
@Beni: Sorry, I intended for the edit to be a second part to the question. –  j_z Mar 17 '12 at 10:58

3 Answers 3

up vote 1 down vote accepted

I will assume that $A=[0,+\infty)$. From definition of limit $$ \lim\limits_{x\to+\infty} f(x)=-1\Longleftrightarrow \forall\varepsilon>0\quad\exists\delta>0\quad\forall x\quad(x>\delta\Longrightarrow|f(x)+1|<\varepsilon) $$ Take $\varepsilon=1/2$ and for respective $\delta>0$ consider point $x_0=2\delta>\delta$, Then you get $f(x_0)<-1+\varepsilon<-1/2$.

Since $f$ is continuous on $[0,+\infty)$ it is continuous on $[0,x_0]$ and moreover $f(0)=1>0$, $f(x_0)<-1/2<0$. Then by Intermediate value theorem there exist $c\in[0,x_0]\subset[0,+\infty)$ such that $$ f(c)=0. $$

On the other hand if $A\subset[0,+\infty)$ then we can construct a counterexample. Indeed consider function $$ f(x)=-\frac{x+1}{x-1} $$ defined on $A=[0,1)\cup(1,+\infty)$ It is easy to check that $$ \lim\limits_{x\to+\infty}f(x)=-1, \qquad f(0)=1 $$ But there is no $c\in[0,1)\cup(1,+\infty)$ such that $f(c)=0$. This is my counterexample.

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From your question it seems that $[0,+\infty)\subset A$. A continuous function $f\colon[0,+\infty)\to\mathbb{R}$ cannot have a vertical asymptote at a point $\alpha\in(0,+\infty)$; such an $\alpha$ would be a point of discontinuity.

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I've edited to include the domain. Does this still apply? –  j_z Mar 17 '12 at 9:00
    
The notation $f:\mathbb{R}\to\mathbb{R}$ implies that the domain is $\mathbb{R}$. So $\alpha$ can't be a real as I suspect you want it to be. –  Patrick Mar 17 '12 at 9:10
    
@Patrick: Thanks, edited again. –  j_z Mar 17 '12 at 9:13
    
Yes, it still applies. A continuous function $f\colon A\subset\mathbb{R}\to\mathbb{R}$ cannot have a vertical asymptote at an interior point of $A$. –  Julián Aguirre Mar 17 '12 at 9:46

It seems the situation you have in mind is when $f$ is defined and continuous on $(0,a)\cup(a,+\infty)$, $f(x)\to+\infty$ when $x\to a^+$ and $f(x)\to-1$ when $x\to+\infty$ (the other hypotheses being irrelevant). Then yes, there exists $x^*$ in $(a,+\infty)$ such that $f(x^*)=0$.

The proof uses the intermediate value theorem and is as follows. First, since $f(x)\to+\infty$ when $x\to a^+$, there exists $x_1$ in $(a,+\infty)$ such that $f(x_1)\geqslant1$. Second, since $f(x)\to-1$ when $x\to+\infty$, there exists $x_2$ in $(x_1,+\infty)$ such that $f(x_2)\leqslant-\frac12$. Now, $f$ is continuous on $[x_1,x_2]$ and $f(x_1)\gt0\gt f(x_2)$, hence by the usual intermediate value theorem there exists $x^*$ in $(x_1,x_2)$ such that $f(x^*)=0$.

Edit: If the hypotheses are modified (drastically) and one now assumes that $f$ is defined and continuous on $[0,a)\cup(a,+\infty)$ with $f(0)=1$, $f(x)\to+\infty$ when $x\to a^-$, $f(x)\to-\infty$ when $x\to a^+$ and $f(x)\to-1$ when $x\to+\infty$, then there is no reason to believe there exists $x^*$ in $[0,a)\cup(a,+\infty)$ such that $f(x^*)=0$. In fact, the behaviour of $f$ on $[0,a)$ and its behaviour on $(a,+\infty)$ are unrelated and, on both intervals, the sign of $f$ may be constant.

For a specific example, consider $f(x)=\frac{a}{a-x}$ for every $0\leqslant x\lt a$ and $f(x)=\frac{x}{a-x}$ for every $x\gt a$.

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I've edited the question again, would this still apply? –  j_z Mar 17 '12 at 10:38
    
My post answers exactly the revised version of your question. Please read. –  Did Mar 17 '12 at 11:44
    
"First, since $f(x) \to +\infty$ when $x\to a^{+}$ there exists $x_1$ in $(a, +\infty)$ such that $f(x_1) > 1$", but in the revised version, is this necessarily true? –  j_z Mar 17 '12 at 11:55

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