A homework question:

I know the solution but I don't know how to prove that the function is negative XOR positive for $x \in \mathbb{R}$

f is continuous in $\mathbb{R}$. $$\text{ prove that if } |f(x)| \ge x \text{ for } x \in \mathbb{R} \text { then: } \lim_{x\to\infty} f(x) = \infty \text{ or } \lim_{x\to\infty} f(x) = -\infty$$

Now once I prove that the function is negative XOR positive it's relatively simple to prove that the limits are at infinity.

How do I prove that there is no $f(x) = 0$?


  • 2
    $\begingroup$ The function f(x)=x satisfies the conditions, and it's neither positive nor negative for all $x \in \mathbb{R}$. Perhaps you need something slightly different? $\endgroup$ – Alon Amit Apr 30 '11 at 15:22
  • $\begingroup$ Increasing/decreasing? $\endgroup$ – The Chaz 2.0 Apr 30 '11 at 15:30
  • 1
    $\begingroup$ Hint: since you want to calculate the limit as $x$ goes to infinity, you might as well only consider $f(x)$ on the interval $[N, \infty)$ for some large number $N$. What does the inequality tells you about the abs value of $f(x)$ on this interval? Can it be zero? Note that nothing can be deduced from the inequality on the interval $(-\infty, 0]$, so you can not say $f(x)$ is positive or negative for all $x \in \mathbb{R}$. $\endgroup$ – Jiangwei Xue Apr 30 '11 at 15:30
  • $\begingroup$ @Cu7I4ss: Suppose for example that $f(x)$ is positive at $x=a$, where $a>0$. Suppose also that $f(b)$ is negative for some $b>a$. Then $f(c)=0$ for some $c$ between $a$ and $b$ (Intermediate Value Theorem). $\endgroup$ – André Nicolas Apr 30 '11 at 17:19

The following is a partial reply to the intended question. It looks to me as if you unfortunately did not quote the question verbatim. And the header is rather misleading.

But it is clear that the original question is something like this. Suppose that $f$ is continuous. Show that if $|f(x)|\ge x$ for all $x$, then $$\lim_{x\to\infty} f(x)=\infty \quad \text{or} \quad \lim_{x\to\infty} f(x)=-\infty$$

You say that you can handle things if you can show that $f(x)$ is either (i) positive for all positive $x$ or (ii) negative for all positive $x$. It is indeed true that $f(x)$ is either positive for all positive $x$ or $f(x)$ is negative for all positive $x$. But to save a tiny amount of time, we will show that either $f(x)$ is positive for all $x \ge 1$ or $f(x)$ is negative for all $x \ge 1$. That is plenty enough to allow pushing the rest of the proof through.

There are two possibilities: (i) $f(1)$ is positive or (ii) $f(1)$ is negative. Note that we cannot have $f(1)=0$, for $|f(1)|\ge 1$ by the condition the problem put on $f$.

We deal with possibility (i). Dealing with possibility (ii) is essentially identical (or cheat: if $f(1)$ is negative, let $g(x)=-f(x)$.)

We will show that if $f(1)$ is positive, then $f(x)$ is positive for all $x >1$. Note that $f(x)$ cannot be $0$ at any $x >1$, for that would violate the condition $|f(x)|>$. We next show that $f(x)$ cannot be negative at any $x>1$.

For suppose to the contrary that $f(b)<0$, where $b>1$. Our function is positive at $1$, and supposedly negative at $b$. Since $f$ is continuous, by the Intermediate Value Theorem we can conclude that $f(c)=0$ for some $c$ between $1$ and $b$. This, as we observed already, is impossible because $|f(c)| \ge c$.

So in case (i), $f(x)>x$ for all $x \ge 1$, from which the statement about the limit as $x \to\infty$ follows easily.

I have written out case (i) in gruesome detail. Of course the argument can be streamlined, but be careful to preserve meaning when you do!

  • 1
    $\begingroup$ An excellent job of mind-reading. I had no idea what Cu7l4ss was trying to ask. $\endgroup$ – Gerry Myerson May 1 '11 at 0:41
  • $\begingroup$ @user6312 I'm sorry for poorly writing the question, but this is how it was written in the handbook. I've done one mistake writing the condition falsly, but nevertheless you and the other folks here answered perfectly thanks! $\endgroup$ – Cu7l4ss May 2 '11 at 4:27

Given the condition on $f$ it is possible that $f(x)=0$ for $x=0$ (and only for $x=0$). So it is not true that $f(x)$ can never be zero. Furthermore, $f(x)=x$ fulfills the requirement and is neither positive nor negative for all $x\in R$.

That the limit is $\pm\infty$ and nothing else you can nevertheless prove (even though the other statements in your post are wrong). For $x>0$, there are only two possibilities: either $f(x)>0$ or $f(x)<0$. Together with $|f(x)|\geq x$ you obtain the statement.


The function, as you stated it, is not exclusively negative nor exclusively positive.

There's a really simple counter example. If f(x)=x for all x in R, then f is continuous everywhere, |f(x)|>=x everywhere, and even one of the limits is satisfied (limit as x goes-to infinity of f(x) is infinity). But clearly, f(0)=0, f(-1)=-1, and f(1)=1. So there is a point where f(x)=0, and f is not exclusively positive nor exclusively negative everywhere.

As to the most likely question you were probably asking, user6312 already answered it for you, but I'll type the same proof for completeness. (Also, does anyone know where I can find a guide on how to get latex to work properly in here? I can't get things like /mathbb{R} or /infinity or /in to work. Maybe the rules have changed since I last used latex...)

If f(1)>=1, then f is positive for all x>1. (Suppose there exists b>1 such that f(b)<1. Since |f(x)|>=x for all x, and b>1, f(b)<-b<0. But if 10 and f(b)<0, there exists a point c, 1the coolest theorem ever. Since c>0 and f(c)=0, =><=)

Since f(x) is positive for all x>1, then |f(x)|=f(x) for x>1. Thus the limit x->infinity of f(x) is greater than limit x->infinity of x which is infinity.

If f(1)<=-1, then let g(x)=-f(x), and do the same proof, and you'll end with lim x->infinity f(x) = -infinity.

  • $\begingroup$ You can use latex when wrapping text with dollar signs: one dollar sign for inline latex and double dollar sign for multiline text And you used the wrong slashes its \ thanks for your answer! $\endgroup$ – Cu7l4ss May 2 '11 at 6:21

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