# Precise definition of limits at infinity

## Homework Statement

Let f be a continuous function on ℝ. Suppose that $\mathop {\lim }\limits_{x \to - \infty } f(x) = 0$ and $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$. Prove that there exists a number M > 0 such that $\left| {f(x)} \right| \le M$ for all $x \in ℝ$.

## Homework Equations

$\mathop {\lim }\limits_{x \to - \infty } f(x) = 0$ ⇔ for every ε > 0 there is N such that if x > N then $\left| {f(x)} \right| < ε$

$\mathop {\lim }\limits_{x \to - \infty } f(x) = 0$ ⇔ for every ε > 0 there is N such that if x < N then $\left| {f(x)} \right| < ε$

## The Attempt at a Solution

I can see something similar to the precise definition of limits at infinity in the question but I'm not sure if this is the case. Any hint is appreciated, thanks a lot!

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jambaugh
Gold Member
You will have to invoke two definitions here. The fact that "f is a continuous function" is vital.

You might start by considering how the graph of f must look given all the "givens". Try to draw a graph which violates the assertion to be proved yet satisfies the given conditions.

Also look at the negation of the assertion to be proved and see if it fits another definition you've recently studied.

You will have to invoke two definitions here. The fact that "f is a continuous function" is vital.

You might start by considering how the graph of f must look given all the "givens". Try to draw a graph which violates the assertion to be proved yet satisfies the given conditions.

Also look at the negation of the assertion to be proved and see if it fits another definition you've recently studied.
Thank you. It seems the existence of such an M is dependent on the continuity of f(x).
Let's take f(x)=1/x, which is not a continuous function. Then the graph of |f(x)| will have both horizontal and vertical asymptotes. In this case such an M doesn't exist.
On the other hand, when I take f(x)=1/(1+x^2) the graph of |f(x)| is nothing but the same as that of f(x). It's easy to pick an M that satisfies the given conditions.
Btw, if I'm not mistaken then the negation of the assertion to be proved is 'For all number M, M<0 and |f(x)|>M'. Also, from what I've learned this far, only the Intermediate Value Theorem and the Extreme Value Theorem require the continuity of the function, but I don't really see any point.

jambaugh
Gold Member
Thank you. It seems the existence of such an M is dependent on the continuity of f(x).
Let's take f(x)=1/x, which is not a continuous function. Then the graph of |f(x)| will have both horizontal and vertical asymptotes. In this case such an M doesn't exist.
On the other hand, when I take f(x)=1/(1+x^2) the graph of |f(x)| is nothing but the same as that of f(x). It's easy to pick an M that satisfies the given conditions.
Btw, if I'm not mistaken then the negation of the assertion to be proved is 'For all number M, M<0 and |f(x)|>M'. Also, from what I've learned this far, only the Intermediate Value Theorem and the Extreme Value Theorem require the continuity of the function, but I don't really see any point.
for all M there exists an x such that ...
You almost have the definition of an infinite limit. But not quite. I'm not saying this is helpful for a proof. It is an observation helpful for understanding the particular situation.

Basically you have to show the function is bounded. Note that the def of (finite) limits at infinity shows it is bounded beyond some sufficiently large values of x. So to be unbounded it would need to be unbounded on a finite interval.